Tire Pressures and Contact Areas - Heavy Equip Tires

[Mazzman]

We’re evaluating the effects of heavy construction equipment operating near the edge of earth slopes. We are attempting to model the vehicle loads and want to consider them as distributed loads per tire over the tire’s contact area. Can we calculate contact area as load divided by tire pressure? Then, we would use the tire’s section width (or tread width) as one dimension of the loaded area and the other dimension will be calculated from the contact area.

The manufacturer of a piece of equipment we’re evaluating uses 29.5R25 tires. The loaded operating weight on one axle (two tires) is 53,270 pounds, so 26,635 pounds per tire. Manufacturer reports the maximum ground pressure on the same axle (loaded) is 22.7 psi.

Based on data I’ve seen from a tire manufacturer for the 29.5R25, the 30-mph load limit at a cold inflation pressure of 29 psi is 16,500 pounds.

So I have:

Equipment Manufacturer: 26,635 pounds produces max ground pressure of 22.7 psi

Tire Manufacturer: 16, 500 pounds is load limit for inflation pressure of 29 psi

If ground pressure and inflation pressure are synonymous, it seems the tire would be overloaded on this equipment.

Suggestions?

 

[CapriRacer]

No, You can NOT calculate the average ground pressure of a tire footprint directly. It is geometry related and you'll have to get someone to give you footprints of typical tires at typical loads and inflation pressure. This is not inconsistent with the data you have been given.

- BUT -

Being engineers also means we can make assumptions and estimates and one of those would be the you could use the formula you suggested with the thoughts that it will not be accurate. I've seen it off by a factor of 3!

 

[bradleyelwood]

Something doesn't sound right. Are you SURE that the tire carries 26.6 kips at only 22 psi inflation pressure? The pressure in the tire is very likely much higher. Are you sure of the the load?

There is a relationship between tire pressure and contact area. If you really want isobars of contact presssure vs load vs pressure, get in touch with the tire manufacturer. If you want an approximation, use an average of contact area times inflation pressure equals load. The published information usually shows the pressure near the center of the contact patch is 1.2 times inflation pressure and diminishes as you move outward.

If we take your 26,635 pounds and divide by 22 psi, we get 1211 square inches and if we divide by maybe 27 inches of width for the tire face, we get 45 inches of length of tire patch. It doesnt sound right.

 

[CapriRacer]

Here's a link that addresses the inflation pressure versus contact patch size that illustrates that there is no connection between the 2:

http://www.performancesimulations.com/fact-or-fiction-tires-1.htm

 

[Mazzman]

To bradleyelwood:

The equipment manufacturer (John Deere) reports the middle axle load (two tires) on an articulated dump truck is 53.3 k, so I assume one tire carries 26.6 k. John Deere data states that the "maximum ground pressure (loaded, middle axle)" is 22.7 psi.

This is NOT a tire inflation pressure.

I went through the same evaluation you report and concluded that if the John Deere equipment data is correct, there must not be a direct relationship between inflation pressure and contact area. CapriRacer is confirming that, and I'll take a look at the link he provides above.

 

[bradleyelwood]

Mazzman,

I stand corrected, the "tire inflation pressure" is 29, your "ground contact pressure is 22.7". But my calculation explains how to go about it:

I assume that you have access to the vehicle and to all kinds of things like rulers, measuring tapes, etc, maybe even a pressure gage. (some off highway trucks use a larger valve stem.)

With the loaded truck sitting still on flat level ground, go measure the tire contact patch, ie the width and length of each tire contact patch. Then measure the tire pressure. Do this for all 6 tires.

Then estimate the total load, this is truck tare weight plus rock or dirt being hauled. (The tare weight of the truck is published.) Then estimate the weight of the contents. If you dont have a way of estimating the dirt, just use 2700 pounds per cubic yard of material. Or, just ask the dirt contractor or the loader operator.

For a quick and dirty method, just use this rule of thumb: A 35 ton truck weighs 35 tons and the LOADED axle weight usually divides equally among the tires. A loaded articulated truck with three axles will have 6 tires, more or less equally loaded. From the little information you have provided, my guess is that you have 35 ton trucks, is this correct?

Now with the above, you will have more than enough tire loading data to make up your own mind on how it works. This information will probably be within 15 per cent. Plenty close enough for soil calculations. Probably not close enough for race cars.

Incidently, my previous comment still stands about "something doesn't seem right" regarding tire pressure for these loads. I looked up Yokohama tire data and at 30 MPH they say the recommended tire loading is 26,000 pounds at 62 psi, or 27,600 pounds at 65 psi.

Let us know how the pressure gage and ruler data turns out.

 

[bradleyelwood]

oops, i wasnt clear about one other thing.

A 35 ton truck hauls 35 tons of payload. The tare weight of the truck is probably about 35 tons. For a loaded three axle articulated truck, the weigh is about 140 kips divided equally to the three axles or about 47 kips per axle, 23 kips per tire.

 

[chicopee]

The footprint of a tire on soil should be close to elliptical in configuration.