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# Calculation of bearing pressure for heavy excavators

## Calculation of bearing pressure for heavy excavators

(OP)
Hi there,

Quick(ISH) question to pick your brains.

I have been looking at bearing pressure for a large item of plant which is on wheels. I've used formula from EC7 based on a shallow foundations. Basically running through the calcs means that the item of plant is no where near acceptable.

Max load on single wheel ~ 85kN and assumed effective bearing area of 0.36m².

I suppose my question is a) is the method for bearig capacity based on shallow foundations reasonable (can't think of/find any other suitable calc and everything seems to add up) and b) is there a method for effective area for wheels, I took area of 0.6m X 0.6m by measuring the thing.

Problem boils down to high load small area but could it reasonable to negate the top few 100mm of soil (accept some settlement etc) and assume a load distribution (say 45-phi/2) and assume load distribution to give a larger effective area?

I've physically seen the machine working no problem but this calc shows I need pads for all soil types up to a phi value of around 42° which doesn't seem quite right to me.

### RE: Calculation of bearing pressure for heavy excavators

If the wheels have pneumatic tires, the usual approach is to assume the soil contact pressure is equal to the air pressure in the tire. Divide the load by the tire pressure to get the bearing area. The footprint is roughly elliptical.

You can approach it as a shallow foundation with no embedment. There has been a lot of study of stresses in pavements, bases and subgrades - See Principles of Pavement Design by Yoder. If you intend to run this on open ground, you might look into trafficabilty principles for military vehicles.

### RE: Calculation of bearing pressure for heavy excavators

Based on the above info, looks like you are looking at about 250 kPa for bearing pressure. Do you have a layer of compacted aggregate base course for traffic purposes? If so, since I guess that you just want to prove with calcs that this is working, assuming a phi of 42 degrees may be appropriate for that base course. For bearing capacity purposes, the aggregate layer does not need to be very thick based on the width of the loaded area (small).

### RE: Calculation of bearing pressure for heavy excavators

(OP)
aeoliantexan: thank you for that, I'd found that relationship and will be taking that going forwards. I'll definitely look into the military vehicles information.

Okiryu: the issue is that the use of the machine is the opposite - the machine is 'marketed' for use on all terrain, in fact if it doesn't work on un-paved ground then the machine is useless.

If I'm being honest when I saw the machine i was surprised as let's face it there's a reason heavy plant is supported by outriggers. The issue is these machines were purchased specifically for use in areas a typical item of plant would not be able to access e.g over rough terrain or poor ground. Since the purchase proof has been requested(and rightly so) that it is workable and my initial views are it's probably not, I've asked on the off chance there is another way of analysing an unsurfaced road as its not my area of expertise by any stretch of the imagination.

Thank you very much for your replies.

### RE: Calculation of bearing pressure for heavy excavators

Are we talking about a crane here?

If it's an "all-terrain" crane-- you're right, those often are not meant for use on unprepared soils. All-terrain cranes are called as such because they can both travel on (good quality or prepared) sites and on public roadways. Whereas a rough-terrain crane can handle softer ground on-site, but would need to be transported between sites.

Either way, RT and AT cranes both often use outriggers when working. (RT cranes sometimes have a much more limited chart available for working "from rubber").

----
The name is a long story -- just call me Lo.

### RE: Calculation of bearing pressure for heavy excavators

Maybe you can run some plate bearing tests? 0.3 and 0.75 dia plates as the lower and upper bounds of the surface support values?

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