Loading on 2nd Floor Concrete Floor

[dmech]

Hello Structural Guru's,

I'm a mechanical engineering and develop automation equipment. I have a situation where our customer has provided a 5 kPa (104 psf) maximum 2nd Floor loading.

The floor thickness is 6" and it is reinforced with rebar. I don't know the details of the reinforcement or the span lengths. Our automation equipment will rest on the 2nd floor at points within the total footprint. The total foot print of the machine is 18.5 m^2 (199.5 ft^2). There are about 30 points loads of varying intensity within the footprint. The maximum point load pressure on one of the equipment footings is 14.00 kPa (292.4 psf). Obviously, this will not pass the 5.0 kPa (104 psf) requirement.

My question is can you assume that the equipment weight is distributed over the total footprint (resulting in a passing design) or must you determine the safety of the loading looking at each point load (resulting in a failed design)?

My gut says the design is okay...based on the simple fact that my shoe is 4" x 12" and I weight 193lbs. Therefore, I would place a 13.84 kPa (289.5 psf). If my gut is wrong my wife and I better not dance on that floor....ever!!!

Thanks in advance for you thoughts and assistance.
dmech

 

[KBVT]

The first think you need to look at is the punching shear at each support point. If you can satisfy that requirement, then you can look into the flexural capacity of the floor.

 

[steellion]

See thread on "Point loads to distributed loads". Converting a point load to a surface load in the manner you described just isn't valid. Your comment about the shoes is a good example. 104 psf is meant as an average for an overall floor load. KBVT is correct. Check the slab for punching shear at the point loads (this is where the surface area does matter), and then there are other checks such as two-way shear, flexural capacity, and deflection to calculate.

 

[dmech]

Stellion, I actually read the "Point loads to distributed loads" thread and then questioned my "shoe" experiment and gut. Hence I posted here.

Anyhow, when going outside my expertise I continually find that "keywords" are a great starting point to learning a subject and thereby further my knowledge. So, thank you both for "punching shear", "two-way shear", and "flexural capacity".

 

[aayjaber]

If you have high concentration of load on a small area where punching force is high then you may consider using some kind of bearing plates of appropriate thickness to reduce the punching force and help distribute the load in a safer way to the floor.

 

[spats]

I think maybe dmech is looking more for a general concept than the technical details. Let me try to explain how I look at it: the floor is designed for 104 pounds placed on every square foot, over the entire floor area. This is not a matter of bearing pressure on the foot of the worst concentrated load, but how much total load is placed in a given area. It's important to know what the span of the concrete slab is. If the slab is supported by parallel beams or joists (concrete or steel)at a somewhat uniform spacing, then it's a lot simpler to explain. In this case you have what's refered to as a one-way slab. The slab spans from beam to beam, and you can establish some fairly straightforward rules. If there are no beams, except maybe at the column lines, then you likely have a two-way slab, and the answer is not as simple.

An example of a one way slab is as follows: say the equipment is 4' wide x 10' long, with the 10' dimension parallel to the intermitent beams, and the total weight is 5,000#. The spacing between beams where the unit is placed is 10' o.c. Further say 40% of the total weight is concentrated in the first 3' of one end. Therefore, worst case, this would result in 0.4 x 5,000#/3' = 667# per foot of slab width. For both shear and bending, if the unit load is is no more than one-half of what the psf load would be, then it doesn't matter if the load is immediately adjacent to a beam, or anywhere in between. Punching is a different matter, but it will not normally control unless you have a very large load on a very small footprint.

In your case, 104 psf x 10' span/2 = 520# per foot of slab width. Therefore, one-half of the psf load is less than the worst actual load placed (pounds per foot of slab width), so it apparently doesn't work. Only a more detailed analysis will give you the real answer. However, the procedure I describe will always be conservative for an otherwise unloaded area. If the 4' width of the unit straddles a beam, then you can look at each slab span individually. If it straddles, and if the unit weight is basically balanced side-to-side, you would only have 333# per foot of slab width, and it would work.