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Point Load on One-Way Slab

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ForrestLowell

ForrestLowell

Structural
Aug 5, 2008
31
I have a floor system with steel beams 6 feet on center and a 12 inch concrete slab on top of them. The steel beems span about 20 feet so the slab acts like a one-way slab. In addition to live and dead loads, during construction I am going to have some pretty large point loads from small mobile cranes that will be driving around to lift equipment into place. How do I treat a point load on a one-way slab system? Everything I can find in the codes and books always assumes that there is a uniform load. But I ahve point loads. What I want to do, and my calculation checker is not agreeing with, is take my point load and distribute it over a width of 4 times the slab thickness. This is what they do in the ACI 318 (13.2.4, 8.10.2) for composite beams, on each side of the web they extend the effective area of flange 4 times the slab thickness of either side. I don't have a composite beam, but I was hoping to use this rationale to justify my effective width. Basically I want to slice my one-way slab into widths 4 feet wide for analysis of point loads. Obviously I will still consider my dead and live loads over the whole 4 foot width. In addition to your opinion any information backing your position would be helpful, so my checker and I can come to a concensus. Thanks for the help! I am just out of college and starting to understand how much I did not learn there.
 
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Using effective width analogy isn't correct in my view. They are dealing with two different things.

AASHTO has a slab distribution formula for concentrated loads placed on bridge decks. Bridge engineers distribute concentrated loads on one-way slabs all the time. They use the AASHTO formula as follows:

The concentrated load is spread over a distance E

E = (4 + 0.06 x S) but not more than 7 feet (or, I would add, the center to center spacing of the posts).
S = span, ft.

Taking a small width of slab, such as 12" or 24" seems irrational as you know the slab will essentially deflect as a whole. And, per Hooke's law, where there's deflection, there's stress. Thus, you know the slab will tend to work as a unit over a wider width than one or two feet.

You should also check punching shear through the slab per ACI 11.12.
 
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I use the following from USD's steel deck catalog for concentrated loads on a one-way floor deck between beams.

While shown for concrete fill on composite deck, I think the methodology is easily adaptable to formed, cast-in-place concrete slabs.
 
 http://files.engineering.com/getfile.aspx?folder=e93d01cb-ec7d-438b-ac05-6c4afdea05b4&file=Di450_0808051256.PDF
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There are several ways to analyze it:
1) finite element model
2) effective width method
3) Westergaard method (Today's AASTHO spec is based upon this) It was originally written in 1930 for the Bureau of Public Roads. The original report is more useful for building engineers than the formulas in the AASTHO spec. which have been specifically rewritten to be used with HS20-44 loadings. I suggest getting a copy of the original H. M. Westergaard report thru your local library. The title is "Computations of Stresses in Bridge Slabs Due to Wheel Loads", Public Roads, 11, No. 1, March 1930.

You will need to analyze for the cranes located parallel and perpendicular to the beams. Include impact for the cranes moving across the slab. Do the cranes have outriggers? If so, those could be even larger than the wheel loads.

You need to check bottom and top main steel and distribution steel. Also check beam and punching shear in the slab.
 
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The USD method is way too conservative for a solid slab formed with plywood! The solid slab has the same stiffness in each direction whereas a slab poured over metal deck has different stiffnesses in each direction.
 
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Some Points of clarification. The slab is supported by metal decking running across my 6 foot span, it is already poured in place so the rebar is fixed. They are #9's at 12" top and bottom and each way. A considerable amount of steel. That E equation from JAE does indeed look to support my thought that it is somewhat bigger than 1 or 2 feet. Do you think that the AASHTO code could apply to an elevated slab? The equation from the steel deck handbook seems to limit me to the width of the wheel plus one slab thickness on either side. I am not liking that one as much, but it might be what I have to use. I am open to more ideas, I am going to try to get my hands on that AASHTO code and see if that is applicable to me. I don't have an AASHTO code readily available as I am working on a building, but I could figure something out. Thanks for all the help.
 
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It sounds like you have form deck and it is important to have a well reinforced slab. I would not use the steel deck method. The thicker the slab the less the % difference is between the slab stiffness (with deck) in each direction.

If you use the effective width method, I would suggest that the effective width for one wheel load does not exceed
0.5xE + 0.5x(the space between wheels).

 
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I would not rely on any of the steel from the form deck as that is probably only 3/4 inch thick, and was intended to support only the weight of the concrete.

Try calculating the maximum resisting moment of the slab with the #9 bars over a 12" width. Then look at the worst moment generated by the point load based on your span condition. See how far you have to span the load out and see if it is reasonable. At 12" thick on a 6' span with #9 bars, this was paobably designed to support four feet of gold at Fort Knox.

Mike McCann
MMC Engineering
 
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All,

Thanks for the help, in the end I am going to use the Westergaard equation that seems to be referenced in several documents of We=0.58*s+2c. Where s is span, c is diamter of applied point load. This comes from Westergaards original analysis in his 1930 publication called "Computation of stresses in bridge slabs due to wheel loads". One of the more experienced engineers in office happened to have access to it. It was a report written for the department of agriculture. The best part is that my checker agrees! Thanks again for all the help!
 
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The effective width that you are using is for one concentrated load only. Don't forget that since you have multiple wheel loads, you will have overlapping stresses. Therefore, you will have to reduce the effective width to take into account the multiple wheel loads.
 
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