Impact Force Help
Impact Force Help
(OP)
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?
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?





RE: Impact Force Help
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.
RE: Impact Force Help
Maybe there is another way around this?
RE: Impact Force Help
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.
RE: Impact Force Help
If you hit to material together which are inelastic then then are going to bounce several times before coming to a dead stop.
RE: Impact Force Help
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.
RE: Impact Force Help
TTFN

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RE: Impact Force Help
Cheers
Greg Locock
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RE: Impact Force Help
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
RE: Impact Force Help
RE: Impact Force Help
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