Oblique travel forces - portal crane

[mte12]

Regarding oblique travel forces for a Portal crane, is it correct to estimate using AS 1418.1 Section 4.6.5?
The capacity of the crane can be up to 500t.

From the standard, the steering force "POT" is proportional to the vertical load, multiplied by the coefficient of frictional contact (KO).
And KO ranges from 0.12 to 0.3.
See attached.
So can the load actually be up to 30% of the vertical load?

AS 1418.1 does say:
“The magnitude of the steering force (POT) depends on the type of crane drives, the crane geometry, and on the coefficient of frictional contact (KO) which is determined by the maximum oblique travel gradient (α).”

 

[mte12]

It looks like that portal/gantry cranes are designed to be flexible so that oblique travel forces are minimised.
Commonly, one of the sides is hinged.

This was in AS 1418.3 Section 3.3.4 Lateral forces on portal cranes:
Lateral forces on portal cranes due to oblique travel may be disregarded where the structure has sufficient inherent flexibility to prevent any effective transfer of such loads through the structure; however, lateral forces due to inertia shall be taken into account.

The other load components, correct me if I'm wrong:
- Laterally, from acceleration due to travel drives. I assume this is from the motor moving the lifted object perpendicular to the rail alignment.
- Longitudinally, this is effectively the friction engaged for travel.

For the load due to travel drives, I'm sure it depends on the configuration, but is there a common range of force values applied, as a % of the vertical load.

Also, I was reading this discussion, somewhat related:
[URL unfurl="true"]https://www.eng-tips.com/viewthread.cfm?qid=449907[/url]
Is there anyway to contact any of these members for an opinion.

 

[GregLocock]

If you are asking about the lateral friction coefficient then car tires can easily give 0.8-1.2, so 0.12-0.3 for a truck like tire seems low if anything.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[mte12]

Hi, no this is for the forces that are transferred to supporting structures.
If anything the 0.3 factor above is probably much too high for these types of cranes.

 

[GregLocock]

OK, so are these wheels on rails or what? i imagine the force is generated when the load is swung along the beam, and starts oscillating.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[mte12]

Yes, on rails. The swinging action is usually termed an oblique force, but in this instance is supposed to be less than what normally occurs for overhead cranes found in buildings.

 

[GregLocock]

So the right hand legs just provide vertical support as there is a hinge at the top of them, so all the axial forces in the beam are reacted by inertia/compliance and steering on the left hand rail. I think it is the height of optimism to say the hinged leg provides steering but not lateral force reaction. I think I'd be looking at Appendix E to see if I can understand their exact thought process, doing dynamics via plug numbers and tables is not something I'm used to, sorry.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[mte12]

By all accounts it would be difficult to come up with anything accurate from first principles.

It is data from tables that I'm looking for. I'm sure there's a good reason why it's not readily available.

Appendix E would not apply for these types of cranes.

 

[GregLocock]

So here's one I did earlier - stability of a gantry as it hits the buffers.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[mte12]

Thanks for your effort but I can't open the file.
The buffer is not what I was looking for though.

 

[GregLocock]

I realise that, but a dynamic model is how I approach these things.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?