Impact Loading 1

[ForrestLowell]

I have a problem. I have an overhead crane that carries 25 tons and moves at 0.5mph. I am incharge of designing the protection so it won't accidently hit something really important. It is designed as basically a jail cell around the important item, what loads should I use when I am designing the the jail cell bars. I think it is a basic impact question. Does i use the full capcaity? 25 tons? more? Less?

Thanks for any opinions or help.

 

[civilperson]

If you allow the protection device to deflect/move a measured amount or be a sacrificial object to be replaced after collision then the static load to apply horizontally shrinks to much less than the weight. Vertical application of the load is the weight plus a impact factor.

 

[kslee1000]

I agree with civilperson's suggestion on designing a flexible housing to obsorb some energy. The following is just my random thought without any aspecific reference - you may estimate the impact force as 10-20% of the lift weight, similar to the suggested longitudinal thrust of crane at the same speed, applied at the point of collision.
Or you can dig out your college books to find solution using energy method.

 

[MiketheEngineer]

Force = 1/2mv^2

Either make it showck absrobing or build it like a tank...

 

[DonPhillips]

There was an article on dynamic forces in the October Structure magazine:

http://www.structuremag.org/article.aspx?articleID=784

Don Phillips

http://worthingtonengineering.com

 

[LobstaEata]

Vertical impact for overhead monorail cranes are typically between 10% and 25%. Horizontal factor can typically be 10% of the vertial design load.

Trolley wheel stops however should be designed to resist momentum forces.

 

[jhardy1]

I would suggest you should really consider such problems using energy considerations, rather than force alone. The kinetic energy of the lifted mass is given by:

E = 1/2 m v^2

This must be absorbed by the impact protection structure as internal work. If repeated impacts are likely, you might want the protection structure to be able to absorb this much kinetic energy as fully recoverable elastic strain energy with no permanent deformation. For a linearly elastic system, the elastic strain energy is given by:

W = 1/2 F U

where F is the maximum reaction force and U is the corresponding displacement. Equate E with W, and you are well on the way to working out whether your protection structure can absorb the kinetic energy of the lifted mass.

If impacts are rare "worst-case" scenarios, you might allow the impact strcuture to undergo some permanent deformation for a full impact. (I.e. design a sacrificial protection strcuture, which may need to be repaired or replaced after a major impact.) In this case, you will find the energy absorption of a well-designed ductile frame can be MUCH higher than its elastic capacity, as large amounts of work are required to permanently deform plastic hinges etc.

Hope this helps!