Converting kinetic energy into a force

[Gart888]

Hi,

I'm currently designing some end stops for a crane, and it has me thinking about some pretty basic questions that you guys may be able to help me with.

I know that the energy that the crane's bumpers absorb is based on the kinetic energy of the crane as it hits the end stop. I know that the force exerted on the end stop is a function of the kinetic energy of the crane, and the deflection in the bumper. The force could also be calculated by F=MA based on how quickly it stops when hitting the end stop.

But what happens if I didn't have a bumper on the crane, and I had steel on steel impact? How would I go about calculating my forces in this case? If the crane actually stopped instantaneously then F=MA would tell me that I had an infinite force between the two pieces. And without knowing the force applied on the end stop, I'm having a hard time knowing how I'd calculate the impact deflection on the end stop (which I'd hopefully use to turn the kinetic energy into a force).

Am I forgetting something simple here?

Thanks!

 

[SnTMan]

See this: For starters...

Regards,

Mike

The problem with sloppy work is that the supply FAR EXCEEDS the demand

 

[rb1957]

in the first case (with the bumper) you have a known stiffness reacting to the impact; a 1D system, "easy" to calc the response.

in the 2nd case (w/o bumper), well, you can't have an instantaneous impact but you do have a very high stiffness structure replacing the low stiffness of the bumper. There are several threads like the above dealing with the force due to an impact event. The key parameter is the time over which the impact happens ... not easy to determine.

another day in paradise, or is paradise one day closer ?

 

[drawoh]

Gart888,

Read up on strain energy. At the exact moment of impact, you have kinetic energy and no strain energy. When your crane reaches a velocity of zero, you have no kinetic energy, and you have an approximately equivalent strain energy. Then it springs back. You will lose energy to damping.

--
JHG

 

[swertel]

I like the link that SnTNMan posted. I will concur with that thread's conclusions. Make an assumption and if you can't live with that result, measure it.

I had a design problem where we were designing a moving assembly line and had to determine if the product and its fixtures would slide or tip off of the moving structure if it came to a sudden stop (like a safety E-stop). We looked at F=ma, impulse, conservation of energy, and spring damping. The end result, we knew the stoppage wasn't instant (infinite force) and therefore guessed at a deceleration time. While I didn't reference MIL-810 like the previous thread, a few milliseconds makes a huge difference relative to infinity.

For another project, that assumption wasn't safe enough. We loaded the mechanism with load cells and accelerometers and ran it into a concrete wall. It didn't tip over or slide, so we verified the design plus got actual data to rerun the analysis.

--Scott
www.aerornd.com

 

[hydtools]

Another impulse = change in momentum problem.
Or work = energy if there is deflection doing work absorbing energy.

Ted

 

[chicopee]

A good crane operator would never intentionally hit bumpers or wheel chocks at the end of the rails. Can you imagine how a suspended load would act in such a case?

 

[GregLocock]

So you don't need bumpers then?

If you have a force hammer and an accelerometer you can spend a happy day measuring the driving point impedance of typical structures. The stiffness is directly related to the low frequency dynamic compliance. A rule of thumb on a car for example is that unless it is a specifically reinforced part you'll get about 5000 N/mm, and if it is a lump of cast iron (engine block) you might see 50000 N/mm. I don't have a feel for what larger structures would give.

Cheers

Greg Locock


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

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[dvd]

Depending on how specific you need to get ... Impact Load Factors

 

[LittleInch]

Gart 888,

One thing you need to think about is that whilst it may seem that there is no movement, everything has some level of compression when hit by a force.

So for your end stop if it was a a large beam it hit, then yes the deceleration might be very fast and the deflection of the beam very small, but it is not zero and hence the force is not infinite.

Also the only energy going into the collision is the energy in the crane at the velocity you choose. even hitting something very heard, the velocity of the crane in the reverse direction will only be slightly less than it went into the collision with.

So the most energy you can get out is the energy you put in minus a bit for noise and heat.

Think of a ball bearing dropping on something.
Drop it on a thick steel floor or beam and it will bounce back to nearly the height you dropped it.
Bounce it on a carpet and it stops dead or bounces a few mm.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[chicopee]

Before you get too involved in the design of end stops, get a copy of a recent American National Standard ASME B30.2 on overhead and gantry cranes. This document will have some details about end stops. The design of end stops is to be predicated on the design of the trolley bumpers, therefore, you'll need the bumper data. The OSHA regulation on cranes is consistent with the standard that I cited.

 

[BUGGAR]

I’ve been designing some crash barriers for local car racers and since there always seems to be questions on this forum regarding design for impact. I suggest the following:

Consider only three variables: the initial velocity, the distance required to arrest this velocity, and the force required to arrest this velocity within this distance (expressed in g’s for later conversion to actual force). The force resisting the velocity may be either linearly elastic, as in hitting a spring; or inelastic (plastic) as in hitting a squishy material that doesn’t spring back.

If linearly elastic, the arresting force in g’s equals the velocity squared divided by the distance times the gravity constant. (v^2) / (dg)

If perfectly plastic, the arresting force in g’s equals the velocity squared divided by the twice the distance times the gravity constant. (v^2) / (2dg)

Practically, It is difficult to achieve a perfectly inelastic constant force resistance mechanism. Crushing beer cans comes close and they can be stacked in parallel to achieve any resisting force, and in series to achieve any distance. If you have enough cans. Drink up.

Usually the arresting force will be a combination of elastic and plastic, and in that case the force vs distance graphs for each will have to be integrated. That will be covered in chapter two.

 

[PNachtwey]

I see there are a few posts that have the right solution or simplified solution
KE = Work
m*v^2/2=integrate(F(x),x,0,?)
work is the integral of force x distance over distance. F(x) doesn't assume the resisting force in linear.
? is the value of x that makes the two sides equal. If F(x) is linear then BUGGAR's equation is correct.



Peter Nachtwey
Delta Computer Systems

http://www.deltamotion.com

http://forum.deltamotion.com/

 

[SreeSiv]

The link posted by SnTMan will get you started.
As discussed in that thread, it depends alot on stiffness of the bodies involved in contact.

Additionally, based on my experience -
Not a good idea to use steel for a bumper structure unless you have a very flexible concept in mind. You would want to use a material that stays in the elastic range during the impact event.
Reason being this -
Steel bumper structure, most probably will see some plastic deformations during the impact. It may cause plastic deformations on the crane structure as well. Impact is displacement based loading (Impact energy = Force x Deformation distance (Elastic + Plastic) ). Which means you will see repeated plastic deformations during every impact. After a few cycles, bumper start seeing fracture.

 

[drawoh]

SreeSiv,

If you know how much energy you have to dissipate, you can design a structure that absorbs it while staying within its elastic range. The structure at the contact point will be springy. Obviously, this is a bad place to engineer in brain-dead mode, and it is not a good job for a CAD[ ]monkey.

--
JHG

 

[BUGGAR]

Skip all of the above and go to a crane bumper/shock absorber manufacturer.