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parking garage live loads

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发表于 2009-9-15 11:28:16 | 显示全部楼层 |阅读模式
parking garage live loads
according the ibc 2006, i need to design for a 40 psf uniform live load or 3000 psf concentrated live load for passenger vehicle parking garage floors.  is it as simple as designing for one or the other?  do i use a combination of the two?  say i have a simple span beam 30' long.  one load case i use the distributed load the other load case i use the point load?  is it that simple?  thanks
you design for 40 psf uniform live load
and
you design for a concentrated load anywhere on the floor.
you must use both - but not in the same load combination.

i agree, you must check both.  i believe the 3k load is for punching shear assuming a heavy suv is changing a tire with a jack.
i have also run into a couple of times where the city requested a lot larger point load to be included in the design in the form of a firetruck.
so what i'm hearing is i design for either the uniform load or the point load but not in combination with one another?
structuraleit, even if a suv is changing its tire on a jack won't the load be still distributed to the 3 tires + one jack? granted the tire's contact area might be a little greater than the jack.
correct ars, "floors in garages or portions of buildings used for the storage of motor vehicles
shall be designed for the uniformly distributed live loads of table 1607.1 orfff">fff"> the following concentrated loads:"
slickdeals-
i think if the jack lifted the suv no higher than the tire did, you would be correct.  in order to take the tire off comfortably, however, you need to jack the vehicle higher than the tire had it which will take load off of some 2 of the other three tires and put it into 1 tire and the jack.  the other 2 tires probably still have some load on them, but much, much less than when all 4 tires are on.
i see nobody here says anything about the surface area the concentrated load of 3 kips must be applied. if designing the slab, then you must consider this load on a surface of 2.5 in by 2.5 in. because of what structuraleit said. the jack. it's really penalizing and is important to be considered.
then the 3 kips load must be considered at each 5 ft. at the same time if designing a beam, if i re  
tlex:
a clarification here...
note a of table 1607.1 of asce7 says the 3200 pound load is applied over an area 4.5" square, not 2.5".
mike mccann
mmc engineering
it's 4.5x4.5 inches and i can't find any mention of applying the 3 kip load at every 5 ft along a beam.
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