huangyhg

dissecting a load combination
from the ibc 2000, formula 16-10 load combination:
d + (lr or s or r) + l + (w or .7e)
given: d = 20 psf, lr = 20 psf, s = 20 psf,
an industrial building with a cab operated overhead crane. when looking at the load combinations that include seismic loads, we take the building weight and we take the weight of the bridge crane mechanism (trolley & bridge, but no lifted load) to figure out the seismic shear force, vertical and horizontal for the building as a whole.
now, look at an individual column supporting the building and the crane (where in this case l = the crane loading on the column; vertical, lateral and transverse).
first, i consider the entire crane load as live load, the trolley, the bridge and the lifted capacity + impact from the single column's point of view for vertical, lateral and transverse loads.
secondly, when considering seismic, it seems that the column sees it's portion of the building seismic load and, when the crane is on the column line in question, all of the bridge crane seismic load (trolley & bridge, but no lifted load) divided between the two columns in the line (assuming a clear span crane bay).
so, the combination for the individual column would look like:
d + .75*(lr + s + l) + .7(e building + e crane) where e crane = total crane mechanism weight used in the seimic base shear equation.
is this how you all look at it too?

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if the column is connected to a diaphragm, then it's seismic load will be distributed to the entire lateral system (the manner in which it is distributed is dependent on whether it is a rigid or flexible diaphragm). if it is a flexible diaphragm, then you appear correct.
the building seismic loading i agree is distributed, but doesn't the seismic load from the crane go almost totally (say 90% to pick a number) directly into the column since it is so close; on the column line?
why would it? it gets dumped into the diaphragm with the rest of the lateral loading. the diaphragm moves as a rigid body which means that all of the lateral frames get affected the same. well, not necessarily the same, but just like a rigid diaphragm.
now, i'm talking just the seismic component, in general a crane runway is some distance below the roof line. so, i'm thinking that the seismic load from the bridge crane weight
"hits" the column first and then is distributed down to the foundation. i guess my question is: do we dump all that seismic load from the bridge crane weight into the one column when the crane is in that column line or into two or three or what? that's what i'm questioning.
i guess i don't know what a crane detail looks like because i've never dealt with one. if it is several feet below the diaphragm, then some of it will go up to the diaphragm and some will go down to the footing, but all of it would put that particular column into bending - from what i'm picturing anyway.
unfortunately, the 2000 ibc does not allow you to multiply 0.7e by the 0.75 factor. this was corrected in the 2006 ibc.
to answer your question, i would put the seismic load from the crane itself into two columns, one each side of the crane bay. when the bridge crane moves laterally in an earthquake, it will distribute its load to both crane runways.

daveatkins
yes, i agree with daveatkins - the columns (the two of them) would see the crane seismic directly.

dave & jae:
thanks, that is what i thought would be the maximum seismic load for an individual column too, the bridge crane stradling the column line and 1/2 of the seismic lateral load in each column at the end of the bridge.
dave, the .75 factor is only applied to the (lr + s + l) not the seismic load or are you saying i'd have to use the full (lr + s + l)? that doesn't see right, i don't interpret ibd 1605.3.1.1 that way.
i think the column next to the trolley would take more e than the column at afar. since the trolley (weight) bears more on it, thus the effect (trolley mass is unevenly distributed in this case).
kslee1000, i don't agree with that. a bridge crane is pretty stiff axially down its length and any lateral thrust from the bridge mass would be distributed 1/2 to each wheel truck and thus 1/2 to each column....assuming the columns have the same lateral stiffness.

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