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What's the best way to calculate tire contact areas? 3
- Thread starter Dezigner
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Did some research and the the way to calc the area is to divide the load per tire by the pressure in the tire...
So if the load per tire is in pounds and you are dividing by pounds per square inch... You will end up with an area in square inches...
Probably, unless of course, the tire is completly flat and resting on the rim...
So if the load per tire is in pounds and you are dividing by pounds per square inch... You will end up with an area in square inches...
Probably, unless of course, the tire is completly flat and resting on the rim...
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The tire contact area is essentially irrelevant if you want to check bearing capacity under a 7 inch slab. The "bridging" effect of the slab is significantly larger than the tire contact area.
Why do you want to check the bearing capacity under the slab? You are not likely to get a bearing capacity failure in confined soil.
Why do you want to check the bearing capacity under the slab? You are not likely to get a bearing capacity failure in confined soil.
Ron's right - the dispersion of the compressive force by a 7" slab will override any "point load" stress - in fact you may be more concerned with an axle load above overlapping boussinesq bulbs - or, if a fill more than 5 feet or so 4 integrated bulbs. That's pretty esoteric for a loader on a 7" slab though - you could just do overlapping 1:4 pyramids starting at tire area plus 7" in all directions for each load.
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- #7
This is actually a fairly interesting problem... The 7" slab is in the bottom of a waste water treatment plant... #6 bars 12" @ o.c. d = 3.5" so the steel is at the neutral axis of the slab... The slab has two spans 10.5 and 12.5 feet resting on cold jointed 12" footings on either side..
The bridging effect of the 3000 psi concrete becomes negligibe when supported by the underlying soil with an allowable bearing capacity of 2500 psf and no effective tensioning strength from the 40 ksi steel...
Also, I'm facing a modulus of vertical sub-grade reaction of 3 tons per cubic foot... Which, if I'm looking at this right, yields a deflection of 0.007 inches or there abouts and craking is a REAL issue with the governing agency...
Further input would be greatly appreciated... I feel that this is a compressional rather than a slab bending problem...
The largest piece of equipment I've been able to prove that will not damage the slab so far is a Cat 963 Track loader with a track pressure of approximately 1500 psf...
The bridging effect of the 3000 psi concrete becomes negligibe when supported by the underlying soil with an allowable bearing capacity of 2500 psf and no effective tensioning strength from the 40 ksi steel...
Also, I'm facing a modulus of vertical sub-grade reaction of 3 tons per cubic foot... Which, if I'm looking at this right, yields a deflection of 0.007 inches or there abouts and craking is a REAL issue with the governing agency...
Further input would be greatly appreciated... I feel that this is a compressional rather than a slab bending problem...
The largest piece of equipment I've been able to prove that will not damage the slab so far is a Cat 963 Track loader with a track pressure of approximately 1500 psf...
Dezigner....I disagree that the bridging effect of 3000 psi concrete over 2500 psf soil is negligible.....I practice in an area where that is the norm and a 7" thick concrete slab is routinely used for pavements carrying all sorts of loads, including those of typical articulated loaders such as the 930.
Further, I can't imagine that your modulus of subgrade reaction is that low in a soil with your stated bearing capacity.
Assuming your 930 loader has a wheel load of say 8000 lb and it has standard tires, you probably have a contact area of about 70 to 80 square inches. Considering this, and the 7-inch slab thickness, the vertical deflection in the soil will be about 0.010 inches directly under the load. At 24 inches from the load, the deflection is about 0.008 inches, thus the load is being bridged (if the load were not being bridged, you would have no deflection that far away from the load and the concrete would have failed in shear). You may compute these stresses and load influence radii with most any convenient pavement analysis program or do it by hand using the PCA method. I did it by elastic layer analysis, considering the concrete and the underlying soil as a 2-layer elastic system. Check Yoder and Witczak for a quick dissertation on elastic layer analysis.
The stress in the concrete at the bottom of the slab directly under the load is about 200 psi in tension. If you are using a concrete with a coarse aggregate of No. 57 stone or larger, you are likely getting a modulus of rupture of greater than about 400 psi, so cracking with a reasonable number of load repetitions is not likely. In fact, in working stress design, we consider that if the actual stresses do not exceed 50 percent of the working stress limit, you will get unlimited cycles of load allowable.
Further, I can't imagine that your modulus of subgrade reaction is that low in a soil with your stated bearing capacity.
Assuming your 930 loader has a wheel load of say 8000 lb and it has standard tires, you probably have a contact area of about 70 to 80 square inches. Considering this, and the 7-inch slab thickness, the vertical deflection in the soil will be about 0.010 inches directly under the load. At 24 inches from the load, the deflection is about 0.008 inches, thus the load is being bridged (if the load were not being bridged, you would have no deflection that far away from the load and the concrete would have failed in shear). You may compute these stresses and load influence radii with most any convenient pavement analysis program or do it by hand using the PCA method. I did it by elastic layer analysis, considering the concrete and the underlying soil as a 2-layer elastic system. Check Yoder and Witczak for a quick dissertation on elastic layer analysis.
The stress in the concrete at the bottom of the slab directly under the load is about 200 psi in tension. If you are using a concrete with a coarse aggregate of No. 57 stone or larger, you are likely getting a modulus of rupture of greater than about 400 psi, so cracking with a reasonable number of load repetitions is not likely. In fact, in working stress design, we consider that if the actual stresses do not exceed 50 percent of the working stress limit, you will get unlimited cycles of load allowable.
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