How Thick Must a Plate Be to Withstand a Projectile Without Failing?

  • Thread starter KetanLobo
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In summary, the question is how thick a steel box needs to be to contain a pressurized projectile of known mass and material properties.
  • #1
KetanLobo
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howdy all

I have come across a strength of materials/mechanics question that I cannot solve.

Basically, the question is..

i have a projectile of known mass and material traveling at a known velocity towards a plate. The plate's dimensions (base and height) are known, as well as the mass and material properties. The thickness of the plate is variable.

My question is... how thick does my plate have to be withstand the force from my projectile and not fail?

i have tried a number of methods with this, and am stumped. I know how to solve this problem if the plate is subject to an applied force, and i can even calculate an impact force on the plate from the projectile. However, i know impact force and applied loads are different, and i don't know how to put an impact force in terms of an applied load.

Also, for a question like this would the shape of the projectile matter?? in this case what's the worst case scenario?

I have access to cosmosworks, and figured i could do a drop test.. but thatll only show me how my projectile deforms, not my plate.

I don't want to shell out tons of money for ansys...

I really am not sure how to solve this, although on the surface it seems like a fairly simple problem..

If you can offer any help it would be greatly appreciated.
 
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  • #2
How is a highly non linear, transient analysis "deceptively simple?"
 
  • #3
ok well that's what i assumed. For some reason it seemed like a question that i'd done in me strength of materials class.

Is there no way to do this without ansys or some other Finite Element analysis program?

Let me make the question simpler for you

We have a pressure testing tube. This tube is pressurized to 5000 PSI. its capped at both ends, one is a threaded cap, the other is welded on. in essence, I am trying to build a containment system around this tube , so that if it pops, or something goes wrong, whatever flies out won't hurt anyone or destroy anything.

The idea, for now, until its feasibility is destroyed, is to make a steel box.

ive calculated the total PE of the air in the tube, and the idea behind my question is to know what thickness i should make my steel box.

ONe idea i had is to simply get steel plating that is rated above 5000 psi. the tube itself is rated to above 5000 psi, so my reasoning is that whatever kinetic force 5000 PSI of my volume of air can generate, as long as the steel wall is rated to 5000 PSI, conservation of energy tells me that as that PE within the air is converted to KE, it won't be able to get through the wall if the wall can take the original PE. If anything, the projectile would lose energy to sound, heat, etc.

However, I'm no mechanics/dynamics/strength of materials genius.

what do you think??
 
  • #4
One possibility is to do a worst-case analysis where all of the potential energy stored in the tube is converted to kinetic energy in the tube cap, and then compare that kinetic energy value to energy values for various caliber bullets. Based on which bullet has the closest energy value, you could then research how thick armor has to be to stop that bullet, and go from there to determine the required thickness of the box.

Alternatively, you could consider surrounding the pressure vessel with sandbags or cinder blocks or other dense energy absorbing materials, but how thick they would have to be I don't know...
 
  • #5
I would start with the question what is the fluid under pressure? If you are compressing air to 5000 psi then you indeed have a bomb on your hands and need to be very worried. If you are pressurizing a fluid like water, which is essentially incompressible, then your concerns are a lot less since you stand a much lesser chance of a catastrophic failure and will more than likely experience a leak.

What is the standard approach is to submerse the item in a water bath. You get some cooling from the compression heating and you get a very inexpensive and effective scatter containment.
 
  • #6
thanks for the input fellas

The pressurized fluid is water. As we fill the tube with h20 we bleed air out. So my worst case scenario is that air isn't bled out and a large amount of air becomes a very small pocket of air at 5000 psi with some fairly high PE.

water submersion isn't a bad idea, but my pressure testing tube is long. ( over 400 inches)

I've also been lookin at ballistic tests but the info is less available than I'd figure.
 
  • #7
you guys know anything about Petry's formula and penetration depth?

I've found some stuff on that but i really don't know if it is applicable.
 
  • #8
I am not familiar with Petry. However, I am trying to find a reference from NASA we use in rotating hardware used for scatter shield thickness. It's not the same thing, but projectile energy is the same. If I find it I will pass it along.
 
  • #10
Very nice...

I found this as well

http://users.rcn.com/harwood.ma.ultranet/t17.html

it seems, however, that I am going to get different results if i do calculations based on either page..

We'll see I'll crunch some numbers today
 
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  • #11
Looks like your link deals specifically with barriers for high-pressure equipment, so I would put more faith in it than the one I posted.
 
  • #12
That seems like a good page to look at. I must admit, this is a difficult problem to accurately predict. That is why we usually just use huge safety factors. Could make a cinderblock enclosure for this testing? I am imagining a cinderblock trench with a steel lid system down the length. It wouldn't be expensive by any means. But I don't know the logistics of your location. There is always the option for an underground trench to test in too.
 
  • #13
im very constricted as to my space, and although a trench is the best idea... someone else told me about it as well, its just plain infeasible for this application. We can't dig up the ground in our warehouse.

ANd this site i found is so so. IF you look in the appendix A you'll see the second equation (velocity at orifice) is pretty convoluted and incorrect. At least i can't get the units to match. It looks like bernoulli and toreccelli's eqs. (v=(2gh)^.5), but where the writer got his version from... i don't know... do you guys know anything about escape velocities of a gas?
the website is almost a decade old as well.

FOr the max velocity my projectile may have I am figuring i take the work done on the gas ( air) as it pressurizes and then setting the work equal to KE to get a velocity.

I assume that since the worst case scenario is that since the air isn't bled its just like the incoming water becomes a huge piston and does x number of joules work on the air in order to get it to 5000 psi.

I used the equation W=-Pf*Vf*ln(Vf/Vi)=KE=1/2mv^2 to get my V.

IS this the most accurate method?

I love the input guys...
 

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