Explosion Containment

[jackboot]

Question:

I have a 15 lb object, and I have a 3/8 thk steel plate.

If my object is a large hex nut with (B7-105 ksi yield) and my plate is A36 (36 ksi yield) supported in a steel frame.

Is there anyway to determine what velocity that the hex nut (or any other object) would need to be in order to penetrate the plate? {Hopefully, I will not have sharp pointed objects in the failure event- but I note that anti-tank ammo is sharp and they cut through some some impressive armor}

The plate will be used to contain a high pressure set-up in the event of a failure. I can think of nothing, equation wise, to work this out.

jackboot

 

[jabonet]

I could compare the penetrating object to some of the existing ammo, I guess a nut hasn`t better penetration that an equivalent mass ammo at the same speed, but it shouldn`t be too far from it, a 50 caliber antitank bullet can penetrate for sure a 3/8 plate.
I`m also sure that there is no way to know it without testing, there are bullets in the market that have quite different penetration, having the same speed and mass, and just different jacket thickness, so who knows, maybe there are some impact test for nuts out there, but who could find them? good luck.

 

[ornerynorsk]

the trick in armored defense is to arrest the mass as much as possible before it hits the failsafe (your plate). Anything that can absorb energy could be packed into a void between the projector and the target: fiberglass wool, steel wool, polyurethane, sorbothane, multi-layer plastics, soil, metal or composite honeycomb, expanded metal mesh, not to mention about a million other things. In addition to arresting velocity, it also spreads the impact force over a larger area. The reason that armor piercing ammo is so effective is because the velocity is very high, and the penetrator material in the tip is non-deforming, it stays in a pointed shape throughout impact, deformation, and penetration. if you're using A36 plate and just want a sure-fire way of stopping fragments during a failure, you may just want to over-build it. A36 is cheap enough. If you have any kind of pressure(and therefore velocity) at all, 3/8 plate is not nearly enough to contain a 15 pound object, regardless of its shape. Good luck!

 

[poetix99]

As "ornerynorsk" has suggested, overbuild it - simply because it is cheap enough to do so, and potentially (very) costly if you don't.

It still begs the question "what constitutes overbuilding in your case?"

If you just want to play around with some numbers, maybe get an order of magnitude sense of what you're dealing with:
- calculate delta enthalpy of the fluid in your "pressure vessel"
(total enthalpy, not specific enthalpy)
- equate that to the kinetic energy of the postulated "projectile"
- equate that to the strain energy of the impacted plate
* use something like the yield (or shear) strength
times the volume of deformed plate
* in effect, solve for the volume of plate required.
I don't suggest that the above be used as a design basis, but I think that you might, as I said, get an order of magnitude sense of things.

 

[arto]

There's an old ASME paper on the subject by R.F.Recht & T.W.Ipson "Ballistic Perforation Dynamics" Trans ASME, Jnl Applied Mech. Sept. 1963 pp384-390.

I'll try typing it here:
Min perforation velocity Vxn
Omega = Plate density/ Projectile density = rhop/rhos
Cp,Cs = acoustic wave velocities in projectile & plate
eta = dynamic shear constant = 1.76x10^5psi in article
phi = [rhop*Cp +rhos*Cs]/[rhop*Cp*rhos*Cs]
T = plate thk
d= projectile dia
L =projectile length
here we go:

Vxn = [4*omega*T^2*phi*eta/(L*d)]{1+sqrt([(L+omega*T)/(omega*T)]*(1+(d/(4*rhos*T*eta*phi^2)))]}

they also give a few other formulas for pointed projectiles & oblique angles

the article should be @ nearest engrg college library

 

[jackboot]

arto,

That looks like what I need.

However, what does the Vxn equate to?

V=velocity?
n=angle(in radians)?

Please let me know.

jackboot

 

[arto]

Vxn = minimum perforation Velocity, normal impact [square to surface]feet per second

C's [speed of sound] are in fps, too

rho's [density] are in (in-sec^2)/in^4

dim's in inches

 

[jackboot]

Thanks very much.

jackboot

 

[Hush]

I would guess you'll never really know without testing it. Thats why they still fire chickens at aircraft windshields and crash test cars. Too many variables, rigidity of the plate, angle of impact, rotation of the object, surface geometry of the object, etc., etc.

I've done a lot of hydrostatic and gas pressure testing and the enclosure is usually made of something flexible, chain mail or some other form of mesh. The suppliers of this stuff should be able to provide you performance data on their product, even if its only - our stuff will stop a 308.

 

[jackboot]

Hush,

You are correct. We have looked at the situation and I have gone and "shot-up" two different grades of steel plate.
Further examination of the failure brought in a host of other problems. For instance, the oil will compress to a point that we will need .6 to .75 gallons extra oil for our test medium. This expansion plus the vessel being tested could potentially release 900,000 ft*lbs of energy.
So, looking at the situation we are going to intentionally weaken a portion of the assembly and make it fail in the test stand. This will be the only way to be sure.
Thanks for the help.

jackboot