Residential Garage slab - suspended

[DKPE]

I am designing a suspended garage slab for a residential home. There was an earlier thread (Mar 2, jjeng2) that asked about the IBC provision of 2000 lb point load vs. 50 psf uniform load. As I read 2000 IBC, Section 1607.4 states that you should compare the 2000 lb point load and the 50 psf uniform load and use the one that produces the greater load effect. The 2000 lb point load always produces a much higher maximum moment on a simply supported structure. When using the 2000 lb pt. load, the slab design seems terribly impractical from the suspended garage floors that I have seen in the field for residential construction. Am I missing something here, or has anyone else run into this problem?

 

[JAE]

How much width of slab are you using to support this 2000 lb load?

 

[DKPE]

I am designing the slab based on 1' "beam" width.

IBC Live load table has footnote that says "2000 lb load acting on an area of 20 square inches."

Section 1607.4 states, "Unless otherwise specified, the indicated concentration shall be assumed to be uniformly distributed over an area 2.5 sq. ft.(6.25 ft2). I assume that the footnote to the table takes precedent over this 2.5 sq. ft. rule. What is your opinion? Even if I can use the 2.5 sq. ft., doesn't that equate to a short uniform load on the 1' beam width of 800 lb/ft over a 2.5 length at the worst location. If so, this still produces a higher moment than the 50 psf load.

 

[JAE]

I read it that Footnote a controls - 20 square inches is the applied area of the load.

But to analyze the slab, the width of slab that resists that load is wider - the slab has to bend in a relatively unform manner. In other words, your little 1 ft. strip of slab cannot bend down, say 1", without the adjoining slab strips also bending down almost the same amount.

If you visualize a one-way slab bending under a concentrated load placed on a 20 sq. in. area, the slab would bend sort of like a curved trough, parallel to the supporting beams, and then slowly bowl upward as you move away from the concentrated loads.

If you remember Hooke's Law - that stress and strain follow each other, then all that bending (the trough - bowl bending) is reflective of moments induced in the slab due to that one load. So that tells you that adjoining strips of slab are also taking some of that load.

But how much slab to use in your calculations? Its tough to know exactly how much without a finite element analysis or using some Roark's Formula (you can if you have the software or book!). AASHTO has formulae for this - they use the formula: Width to use = 4 + 0.06S where S is the span of the slab.

So you would take a strip of slab = 4 + 0.06S to resist your concentrated wheel load plus the dead load of that width of slab.

 

[jike]

If you are using concrete over metal deck, refer to the following:

http://www.njb-united.com/usd/ddds/data33.pdf

 

[theonlynamenottaken]

I posted a very similar thread that had some good responses...

http://www.eng-tips.com/viewthread.cfm?qid=112794

 

[DKPE]

Thank you for the responses. I did research the AASHTO formula and that does help "equalize" the 2000 lb point load vs. the 50 psf uniform load.

 

[Stress02]

There are good references in the USD (United Steel Deck) product manuals about spreading concentrated loads in a slab.

http://www.njb-united.com/usd/dpdfs/dp1819.pdf