ground bearing pressures

[Bert2]

Hi,

i'm trying to calculate the ground bearing pressure for a crawler type crane, given a load at a radius.

attached is an example to start with.

the way im thinking is this;

the load will give a moment at the point which the radius starts, which is of centre from the centre line of the tracks.

then taking the moment as two component forces eg x,y gives a vector resultant force z.

now with z make two components by using cosine etc..

then with both vertical forces ('x') gives you load over the tracks.

then i still need to find out how to calculate the pressure decrease over the tracks to the rear (op to load end)

am i on the right thought path??

just like to add this is theoretical and has no practical / real life implications.

thanks.

 

[Bert2]

anyone?

 

[jgailla]

Bert2,
You haven't gotten any replies, so I'll have a go at it. I've never done crane calcs.
The downward force at the back of the crane is dependent on the stiffness of the soil.
The stiffer the soil, the more load will be spread along the crane tracks.
If you are on a very soft soil, almost no load will be transferred to the rear and the crane will tip forward.
If you are on a very stiff soil, the load will be almost completely transferred along the entire length of the crane tracks, assuming the crane body is a rigid mass.
The actual load on the soil would be somewhere in the middle. There is no way to calculate the load at the back without knowing the strength of the subgrade.

 

[IDS]

The downward force at the back of the crane is dependent on the stiffness of the soil

The pressure distribution is affected by the relative stiffness of the tracks and the soil, but the position of the centroid of the reaction force must be exactly under the centre of mass of the applied load (combined crane and load).

It is a reasonable assumption that the soil pressure is trapezoidal along the track if the centroid of the applied load is within the middle third, or triangular if not.

Using this assumption, the pressure at either end for a trapezoidal distribution (assuming the load is central between the two tracks) is:

W/LB * (1 +- 6E/L)

where:
W = applied load
L = Length of tracks
B = Total track width
E = longitudinal eccentricity of the load centroid from the centre of the tracks.

If E > L/6 then the distribution is triangular and the maximum pressure is:

2W/(3*B(L/2-E))

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/