Impact Force Help

[mymagicpie]

Hi,

This is a simple question (I think). I need some help to calculate an impact force. The information I have:

* The object has a mass of 0.2kg.
* The velocity at the point of impact is 2.5m/s.
* At the point of impact, the object is being accelerated only by gravity.

Using:

Kinetic Energy = 0.5 x mass x velocity^2 = 0.5 x 0.2 x 2.5^2 = 0.625 Joule.

Now to convert this into a force I need to ue the Work-Energy Principle? But to use this I need to know the amount the object would travel after the collision. Which I do not know. The object and its collision surface are both solid steels, hence I would consider this as two inelastic bodies. But this would give me an infinitely high level of impact force. When faced with such a question is there a way to calculate a force?

 

[desertfox]

Hi

Without either the time period the collision lasts or deformation of the materials involved I doubt you can obtain a force.
Can you tell us more bout the application relating to the problem.

 

[mymagicpie]

It is a very simple mechanism. Which involves a mass running down a guided track. The mass is stopped when it collides with the 'stopper' part. The stopper is a raised boss which sits on the track at the end of the stroke. I am simply trying to determine if this boss part is suitably sized so that it does not fail. By fail I mean completely shears off due to the impact of the moving mass. The 'stopper' part can deform, even yield but it has to stop the moving part.

Maybe there is another way around this?

 

[3DDave]

The good news is that no material is infinitely rigid, which prevents loads from being infinitely high. You need to determine what force the moving part can survive and then design the stopper part to be springy enough so that when it absorbs the energy the force on it is smaller than that load.

E = mgh or 1/2 mv^2 = 1/2 kx^2 ; kx = F; so decide on F, then pick an x, then decide if you can get enough k to take the energy.

If the spring is completely elastic, the item will rebound with the same speed it had to begin with. So, you might also want to add a damper to absorb the energy as well. Look into dash pots as a way to slow the rebound. They are used in door closers to keep doors from slamming shut.

Many materials are not infinitely springy - they can be deformed and stay that way. Lead, for example is used to make non-rebound hammers aka dead-blow hammers. They dissipate energy by permanent deformation. Higher energies result in higher loads/more deformation. Other soft metals could also be used as impact targets.

Are you working on stopping a train before it climbs the stairs at an airport? And it was stairs - the news reports of it climbing an escalator were incorrect. Those escalators aren't wide enough for train.

 

[desertfox]

Is it possible to reduce the impact with a spring and damper system?
If you hit to material together which are inelastic then then are going to bounce several times before coming to a dead stop.

 

[mymagicpie]

Thanks, some good suggestions there. However, I am fixed with having a steel to steel contact. Not ideal.

I have worked backwards and found that the 'stopper' will require 5kN load to fail in shear.

So the question is, will the moving mass impact force exceed the 5kN. Engineering judgement would suggest no. But I still need to demonstrate this.

 

[IRstuff]

This subject has been discussed ad nauseum in these forums. I suggest you do a search. The most popular approach, I think, was to use the modulus of elasticity and treat the metal as a spring system.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[GregLocock]

http://www.eng-tips.com/faqs.cfm?fid=1245

Cheers

Greg Locock


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

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[rb1957]

engineering judgement says you're ok ... then go ahead and test it ... even mock something up

since you know your stop is good for 5kN, then figure out the minimum time interval to show it good ...
0.625 = 5000*t ... t = 0.000125sec
maybe 0.625 = Ft/2 is better ... tmin = 0.00025 sec
but this is just playing with numbers ... test it

Quando Omni Flunkus Moritati

 

[Tmoose]

If the 0.2 kg mass is a slightly oversized eagle egg and makes contact with the steel stopper I don't think it will survive the stop in good condition.

 

[hendersdc]

From Roark's 7th addition section 16.4:

For Vertical Impact:

di/d = Sigmai/Sigma = 1 + sqrt(1 + 2*h/d)

For Horizontal Impact:

di/d = Sigmai/Sigma = sqrt(v^2/(g*d))

Where:

di = deformation due to impact
d = deformation due to static load
Sigmai = stress due to impact
Sigma = stress due to static load
h = drop height
v = velocity at impact
g = acceleration due to gravity

These formulas basically give you a way to relate the results of a static FEA or hand calcs to the force and deformation that may be seen during impact. I have applied these formulas in the past to model impact of a round object on a spherical plastic lens for a small handheld device and found them to be fairly accurate if not a little conservative. In my case I did a static FEA of the impactor sitting on the lens with the only force being its own weight. Using the deformation and stresses found from the FEA I applied the formulas using the test drop height and found the deformation and stresses due to impact. The lens was protecting an LCD. During testing the impactor was dropped from increasing heights until failure which is where my opinion of the formulas being a little conservative comes from though I can't remember exact values.

Doug