Particle Impact
Particle Impact
(OP)
Hello. This is my first time on this site. I am hoping to utilize the experience on here. I have a scenario that I have tried to calculate, but I am not sure of my results. Here is what I have.
I have a material with the weight of .305 lbs/ in^2. Size is 25um. For calculation purposes, consider this the diameter of a sphere. The ultimate tensile strength is 144 ksi. This material will be at a velocity of 750 m/s hitting a surface with a tensile strength of 13,500 psi. The material hitting the surface will completely be deformed as needed. The surface it is hitting can not be considered to have a spring factor due to multiple particles at that velocity hitting it. There for my assumption will be no time to spring back. The force I calculate seems to be a bit higher than I would think it should be.
Thank you for your time.
I have a material with the weight of .305 lbs/ in^2. Size is 25um. For calculation purposes, consider this the diameter of a sphere. The ultimate tensile strength is 144 ksi. This material will be at a velocity of 750 m/s hitting a surface with a tensile strength of 13,500 psi. The material hitting the surface will completely be deformed as needed. The surface it is hitting can not be considered to have a spring factor due to multiple particles at that velocity hitting it. There for my assumption will be no time to spring back. The force I calculate seems to be a bit higher than I would think it should be.
Thank you for your time.





RE: Particle Impact
If it is a plastic event then the force will be roughly the momentum divided by the time of collision, and the time of collision will be about 0.5*speed/diameter.
Cheers
Greg Locock
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RE: Particle Impact
you should also define the particle material ... is it undeformed by the impact (ie unobtainium) ? how much energy does it have at the start of the impact ? the next key parameter is the duration of the impact; greg gives you a nice typical number, but play with the time and you'll see it has a significant affect of the results.
"The material hitting the surface will completely be deformed as needed." "There for my assumption will be no time to spring back." so the energy of the impact is absorbed by the particle, as though it has hit a brick (unmoveable, rigid) wall ? i think you're saying that the KE of the particle is transformed into an inertial deceleration of the particle as the impact event happens (ie over the time of the impact the particle's speed is reduced to zero) giving you an impact force.
Quando Omni Flunkus Moritati
RE: Particle Impact
I have a small particle that is completely deforming. In technical terms, the particle is going "splat". :)
I have converted all numbers over into the same system. I was just giving the raw data that I used. I also am looking for the correct formula(s) to see if I made errors.
I used 750 m/s as the delta V being that the speed is reduced to zero. The surface would be unmovable, unless not capable of with standing the force there for deforming itself. In which case the surface is considered to have failed.
A close material to this would be Inconel 625
RE: Particle Impact
Quando Omni Flunkus Moritati
RE: Particle Impact
RE: Particle Impact
thinking about it 750 m/s is well supersonic ... > M2 ?
back to consistent units ... you're using an imperial weight density (lbf/in3). i think your particle has a volume of 6.545E-14m^3, a weight of 1.2E-9 lbs = 2.65E-9 kgf, a mass of 2.7E-10 kg, an energy of 7.6E-5 J (no?)
Quando Omni Flunkus Moritati
RE: Particle Impact
volume = 6.545E-14m3 = 4E-9in3
weight = 1.2E-9 lbf = 5.5E-10 kgf (1kg = 2.2 lbf)
mass = 5.5E-11 kg
energy = 1.6E-5 J
Quando Omni Flunkus Moritati
RE: Particle Impact
25 um diameter = 9.8425 x 10^-4 diameter in inches
4/3 * pi * r^3 = volume of 4.992 * 10^-10 in^3
Weight = 1.522 * 10^-10 lbs = 6.921 * 10^-11 kg
Am I correct, or did I get a different number in there
RE: Particle Impact
V = 5E-10in3 ... wt = 7E-11kgf, mass = 7E-12kg, energy = 2E-6J
having the particle hit a rigid surface gives you a (very) conservative impact force ... do you see a steel particle smearing itself over the plate ?
Quando Omni Flunkus Moritati
RE: Particle Impact
25.4 mm to an 1 inch.
RE: Particle Impact
RE: Particle Impact
Quando Omni Flunkus Moritati
RE: Particle Impact
RE: Particle Impact
Quando Omni Flunkus Moritati
RE: Particle Impact
http://files.engineering.com/getfile.aspx?folder=3...
so
mass = 6.9E-11 kg
vol = 4.99E-10 in^3
energy = 1.94E-5 J
TTFN

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RE: Particle Impact
Quando Omni Flunkus Moritati
RE: Particle Impact
TTFN

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RE: Particle Impact
Quando Omni Flunkus Moritati
RE: Particle Impact
Quando Omni Flunkus Moritati
RE: Particle Impact
Furthermore, we've pretty much narrowed the material to steel, which has density around 8 gm/cm^3, which is what you get when you convert the 0.305 lb/in^2 to metric by multiplying by 453.6 gm/lb and dividing by (2.54 cm/in)^3
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RE: Particle Impact
1lbf = 1slug*1ft/sec2 = 1lbm*32.2ft/sec2
1kgf = 1kg*9.8m/sec2 = 9.8N
"453.6 gm/lb" ... this is a force conversion ... 1kgf = 2.2lbf (a 1kg bag of sugar is pretty much equal to a 2lb bag of sugar).
"when I get my "weight" of 160 lb on my bathroom scale is telling me that my mass is 72.6 kg" ... no, it's telling you your weight is 72.6kgf; scales don't measure mass, they measure weight (force).
steel weighs 0.3lbf/in3.
Quando Omni Flunkus Moritati
RE: Particle Impact
"1lbf = 1slug*1ft/sec2 = 1lbm*32.2ft/sec2"
Exactly as I said. All scales report 1 lb force, which is the force exerted by a 1 lb mass. Therefore, under colloquial usage, something that weighs "1 lb" has mass of 1 lbm
"sorry but you won't convince me that something that weighs 1lb has a mass of 1lb"
Your expression above does exactly that. 1 lbm (mass) --> 1 lbf (force)
"steel weighs 0.3lbf/in3."
By your own conversion above, steel's mass density is 0.3 lbm/in^3.
TTFN

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RE: Particle Impact
Here you have a link to a page regarding consistent units: http://femci.gsfc.nasa.gov/units/index.html
As far as I know kgf (kilogram force) is not a unit in the SI system. In some cases a unit called kilopond was used in that context but that was ages ago. Today it is Newton that we (at least myself) use in the SI system. That does not mean that kgf can't be used but you have to be careful, as always, with the units.
Regards
Thomas
RE: Particle Impact
V = 5E-10in3
wt = 1.5E-10lbf = 7E-10N
mass = 7E-10kg ... 100x my original number ! ('cause i divided kgf by 9.8 instead of multiplying)
or is it force = mass*accel ... mass = weight/g = 7E-11 ... sigh
@IR,
"Slug is the force unit in US units, just like Newton is the force unit in SI."
nope, lbf is the imperial force unit, slug is the native mass unit (mass moments of inertia in slug*ft2) ...
1 lbf = 1slug*1ft/sec2 = 1lbm*32.17ft/sec2
1 N = 1kg*1m/sec
Quando Omni Flunkus Moritati
RE: Particle Impact
I will need to determine the amount of distance the particle will travel into the surface to ultimately equal the amount of compression strength the surface has. Meaning, the particle will indeed be embedded into the surface and the key will be how far. Too far being failure.
Or am I off here?
RE: Particle Impact
something that weighs 1kgf has a mass of 1kg, something that weighs 1lbf has a mass of 1 lbm.
particle weighs 7E-11kgf, mass = 7E-11kg, energy = 2E-5J
(still wonder about the 750m/s though)
"the amount of distance the particle will travel into the surface to ultimately equal the amount of compression strength the surface has" ... i don't think so; i think the KE of the particle is transformed into strain energy of the plate (and particle).
since this is an FE forum, i'd've thought that you'd be doing an impact analysis with MARC or similar.
Quando Omni Flunkus Moritati
RE: Particle Impact
Quando Omni Flunkus Moritati
RE: Particle Impact
You're right, my bad; hopefully, that's my only senior moment this week. At least, Mathcad didn't let me down.
TTFN

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RE: Particle Impact
Imagine a small lead pellet hitting a 12 inch thick case hardened steel plate. The end result is a thinnish disc of lead and no change to the plate.
No slowly increase the yield stress of the pellet, and decrease that of the plate. At some point you will start to see plastic behaviour of the plate. Hertz's formula will give you a good idea for that. Up until that point most of the energy in the collision is taken in plastic deformation of the pellet.
Cheers
Greg Locock
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