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Particle Impact

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.

RE: Particle Impact

I'd start by getting some consistent units, since your density seems a bit odd and the particle diameter is tiny. Also I find it hard to believe that Young's modulus isn't needed, unless you are talking about a completely plastic event.

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

he's got a steel particle ... but consistent units reduce to chance of a simple mistake.

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

(OP)
rb1957 is pretty close.

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

the surface would need to be Really rigid to behave the way you're thinking. your approach is definitely conservative, the surface would certainly dent (is if it was being shot peened) and that'd absorb a lot of the enregy of the collision.

Quando Omni Flunkus Moritati

RE: Particle Impact

(OP)
You are correct. The surface would dent some. As the material builds up, the rigidity would increase as well. There is a lot of energy to absorb. The tricky part is with one particle after another hitting. This is a cold spray method. 750 m/s forming a metal layer. Guess I would have to figure out the amount of or depth of the indentation by the particle into the material to determine the total distance/ time for deceleration purposes. The question becomes how far of an indent along with a particle behind it.

RE: Particle Impact

firing particles along the same trajectory would mean dent on top of dent, strain hardening, ...

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

ok conversion issues ...

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

(OP)
Volume=
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

arh? 25um = 0.025mm = 0.001" !
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

(OP)
25um = .025mm = .025mm/25.4 = 9.8425 x 10^-4 in

25.4 mm to an 1 inch.

RE: Particle Impact

(OP)
I am hoping the particle will smear itself. When plastering it to steel, it works.

RE: Particle Impact

yes, i was rounding

Quando Omni Flunkus Moritati

RE: Particle Impact

(OP)
Sorry, your ARH? Seemed to question where i got my numbers from.

RE: Particle Impact

no, that was me venting at myself, 'cause i took a very round-about way to figure out your diameter as 0.001"

Quando Omni Flunkus Moritati

RE: Particle Impact

i think the denisty is a weight density, not a mass density ...

Quando Omni Flunkus Moritati

RE: Particle Impact

sorry but weight is a force ... you can't use weight to calculate KE (unless you divide by g, in which case you're using mass)

Quando Omni Flunkus Moritati

RE: Particle Impact

i think that's why your number are an order higher than mine

Quando Omni Flunkus Moritati

RE: Particle Impact

I do recognize that that weight is a force. Nonetheless, when I get my "weight" of 160 lb on my bathroom scale is telling me that my mass is 72.6 kg, and not something else. Moreover, the US unit for weight as a force is a slug, i.e., 1 lb * 32 ft/s^2, which is substantially different.

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

TTFN
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RE: Particle Impact

sorry but you won't convince me that something that weighs 1lb has a mass of 1lb. it has a mass (as you rightly say) of 1 slug ...

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

Slug is the force unit in US units, just like Newton is the force unit in SI.

"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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Need help writing a question or understanding a reply? forum1529: Translation Assistance for Engineers

RE: Particle Impact

Hi

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

why did i confuse things with kgf? i guess it seemed like a good idea at the time ...

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

(OP)
So to figure this out....
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

ok, I'm stupid !

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

of course though, doing FE on something so small raises questions about the material of the plate. with a particle of 0.001" diameter grain boundaries are going to be important, probably the assumption of homogeneous material isn't valid.

Quando Omni Flunkus Moritati

RE: Particle Impact

Back to the fundaments of the issue

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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