Particle Impact

[dagger440]

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.

 

[GregLocock]

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


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

http://eng-tips.com/market.cfm?

 

[rb1957]

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

 

[dagger440]

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

 

[rb1957]

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

 

[dagger440]

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.

 

[rb1957]

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

 

[rb1957]

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

 

[dagger440]

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

 

[rb1957]

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

 

[dagger440]

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

25.4 mm to an 1 inch.

 

[dagger440]

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

 

[rb1957]

yes, i was rounding

Quando Omni Flunkus Moritati

 

[dagger440]

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

 

[rb1957]

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

 

[IRstuff]

Ha, another reason to use Mathcad ;-)

http://files.engineering.com/getfile.aspx?folder=3fc94c08-2ff3-4500-9053-b9537b18cc12&file=splat.gif

so

mass = 6.9E-11 kg
vol = 4.99E-10 in^3
energy = 1.94E-5 J




TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[rb1957]

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

Quando Omni Flunkus Moritati

 

[IRstuff]

In the US, weight IS mass, i.e., a .305 lb "weight" is the force produced by a .035 lb mass. That's why Mathcad has a special unit called "lbf" a "pound-force." Note that SI is not immune to such shenanigans, Mathcad also has "kgf" a "kilogram-force."

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[rb1957]

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

 

[rb1957]

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

Quando Omni Flunkus Moritati