Plain Concrete Slab-on-Grade Design

[PT999]

I am designing a slab on grade for a Lifting Mechanism to Support Buses (4 to 6 separate lifting units are used to lift up each bus)

Previous thread on this subject suggested I get the codes for slab design. So I have them now, both ACI Design of Slabs on Grade and PCA design (which are the same).

These codes use charts to determine slab depth based upon among other things the tire contact areas.

My lifting machine has three small wheels. Contact areas for this machine are much smaller than the design charts allow for. So in essence I can't use the charts.

So I am considering ACI 318 plain concrete bearing stress

Nominal Bearing Load =0.85 * fc' *Area *2 ).

(2 because the supporting surface is is wider on all sides that the loaded surface)

Well this lifting machine has three little wheels, set up like a trycle with a total load of 6.75 kips per machine

Spacing of these wheels is 17"

How much load can I put on one wheel? My guess is 2/3 of the total, but it is a little bit indeterminate.

Shear stress is also defined in ACI 318.

So what else should I be checking for. I don't see flexure because it is a slab on grade, although there must be some.

If anyone is really knowledgable on this and wants to get involved in this for a small fee, call me. I also have to work this out for concrete with wwf and bar reinforcement.

Peter 516 681-9239

 

[DaveAtkins]

You WILL get bending in the slab. You see, the soil pushes up on the slab across a greater area than the small wheel contact area, causing bending in the slab. However, if the wheel load does not punch through the slab, and does not overstress the soil in an area equal to the wheel contact area flared out at a 45 degree angle through the slab, then you should not get settlement below the wheel. You will just get slab cracking due to flexure.

DaveAtkins

 

[Ron]

Yes, you will get bending as DaveAtkins said.

This can be designed as a pavement using the PCA technique and the smallest contact area (check your charts...not much difference at small contact side) as it is a good idea on something like this to compute the radius of relative stiffness.

You can also use an elastic layer analysis program (ELSYM5 or similar). This will allow you to input any contact area and determine the bending stresses at the bottom of the slab.

Having said all that, you are probably looking at a 4-inch to 5-inch thick slab (certainly no more). Be careful with thickness control and jointing of the slab. Will make all the difference in performance and crack control.

 

[PT999]

Thanks DaveAtkins and Ron

My contact area is .8 square inches for one wheel. Way off the starting point of 25 sq inches. So I don't see how I can use it.

Your method for analysis sounds good

Bearing Stength of concrete is shown in ACI Structural Section (Sect 10.17) on a 1 on 2 plane instead of the 1 on 1 as you suggested.

Can I use this, but the code limitation is 2 times area from top, or is this not a good analogy for the load on the soil. Naturally this is less conservative, but with 1 on 1 the depth is 11 inches for 2/3 of 6750 lbs, and soil bearing set at 4000 psf.

Also, what am I going to do about the wheel load on this tricycle arrangement. I said before I was thinking 2/3 of total. Perhaps its too much.

 

[tincan]

0.8 in^2 seems awfully small for a wheel contact surface. What are the wheel dimensions? Also, pneumatic or hard rubber tires?

 

[PT999]

This is machine that can lift one wheel of a bus.

To move the machine around, it has 3 small nylon wheels. The wheel contact area is about 2" by 0.4".

Wheel dimensions are about 2" wide x 5" diameter


I think that the concrete fails in bearing, which is only a function of concrete strength (fc'), not thickness

For fc' = 4000: Bearing Capacity = 5586 lbs

Bearing strength factor x 2 x .85 x fc' x wheel contact area
(from ACI code, bearing factor = 0.55 for plain concrete)

Factored Load = 1.6 x 6750 = 10800 lbs, thus NG

10800/5586 or overstress by a factor of 1.9

The manufacturer also says the wheels are applying about 8000 psi to the floor. I haven't looked at the problem from this data, but it strikes me as alot of stress. Looking at it quickly from the now out of date working stress design:

Allowable bearing stress =.25 x 4000 = 1000 psi and the load applies 8000 psi to the floor.
Thus bearing is overstressed by a factor of 8

 

[PSlem]

Go to the BRAB report on slabs.

http://www.nap.edu/catalog/9804.html

I recall there is something in there about any point loads smaller than the thickness of the slab can be considered larger and a walk through of design points.

 

[Ron]

PT999....your factored load per wheel is 1.6(6.75/3)=3.6 kips, not 6.75 kips as noted. Otherwise you can treat the three wheels as acting like one wheel, but your contact area is tremendously increased.

Something seems a bit out of whack in your analysis. Floors in garages get used every day for high point loads from lifting jacks for trucks, buses, and other heavy equipment. They don't fail from this, so I think some of your premises need adjustment.

 

[PT999]

I have had another look at the load criteria, and 6750 is the load per wheel, which matches the 8000 psi stress the manufacturer states.

The machine has an overall lifting capacity of 16000 lbs.

So using 1.6 x 6750, the factored stress is 10,800

Looking at the ultimate stress of 3067 (earlier calc was done wrong) it is way overstressed.

Anyone out there want to look at this for a fee?

 

[apetr26542]

You might want to check out ACI 360R-92 Design of Slabs on Grade Reported by committee 360. You will need to know the modulous of subgrade, a coefficient used in the design of slabs. In order to obtain this you will need a Geotechnical Engineer to perform testing to determine this.