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mathbl2 (Mechanical) (OP)
29 Nov 07 14:28
I need to estimate the ultimate pressure inducted to the ground by a Rought terrain crane on outriggers. This is required for the design a working area .

The crane weigth 79k kg and the component to lift weight 17k kg. Each outrigger (4 total) are 714mm diameter round.

On a theorical base (weight equally separated in 4): (79000 + 17000) * 9.81 / (4 * (0.714/2)^2)*PI)) = 588 kPa.

This pressure look already way too much. Can somebody tell me what i'm doing wrong ?

Thanks
Helpful Member!  minerk (Mechanical)
29 Nov 07 15:40
Well there's nothing wrong with your math, that is the correct number if you assume that the combined weight is evenly distributed over all four outriggers.  I hope you realize that is not a good assumption.  You have to consider the moment imposed by the load.

If you know what make and model of crane you're going to use you might be able to obtain data from the operator's manual regarding outrigger loading for various load/radius/machine setup scenarios.
mathbl2 (Mechanical) (OP)
29 Nov 07 15:49
Thank you for your reply minerk.

I found from the manufacturer (Groove) a chart with the maximum pressure possible on one outrigger.

He's specifiying 1845 kPa, I was with the impression that the crane was not inducting that much pressure.

I will have to review lifting procedure and request an increase of the supported area (crane mat under the outriggers)

Does anyone knows how to have a idea of the ground pressure when the crane is moving. I know it's relation of the tire pressure and the area but I need a rough estimate. Tires are 33.25 x 29. What would be the grounded area ?

Thanks for your post.
minerk (Mechanical)
29 Nov 07 16:57
Rough estimate:  Ground pressure = tire inflation pressure.

The tire will deform until the contact area is large enough to equalize the pressure inside and outside the tire.  You can go to the tire mfg, find the contact area for your tire, divide the weight of the machine by the total tire contact area. The result will be about the same as the tire inflation pressure.  

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