×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# ground bearing pressures

## ground bearing pressures

(OP)
Hi,

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

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.

(OP)
anyone?

### RE: ground bearing pressures

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.

### RE: ground bearing pressures

#### Quote:

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:
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/

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!