Projectile Blast Containment Calculations

  1. Hello all.

    I am working on a assignment in which I have to determine if a projectile travelling at a known velocity will break through a side of its containment room (assuming worst case senerios). I have been looking online and in my textbooks and have ended up confusing myself on what equations I need to use. I have looked at stress equations, THOR equations and piercing equations.

    Can any one please point me in the right direction on what the best way of modelling this problem would be and the best equations to use.

    For information purposes if it makes a difference, the projectile is going to be travelling at a maximum velocity of 84m/s and weights 1kg and the containment room is made out of mild steel currently 1.2mm thick.

    Thank you for any help.
  2. jcsd
  3. SteamKing

    SteamKing 8,626
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    1.2 mm blast containment ain't much containment. You might as well use a beer can.
  4. Yes I think a beer can would be about as much protection, but I have to prove that this is the case so I can then design a better containment room.
  5. You might want to take a look at work being done on asteroid impacts.
    I saw some experiments based on earth collision that probably resulted
    in mathematical modeling that would solve your problem in one go.
    Whether you can find the models and be allowed access I don't know though.
    Start with yahoo or somewhere and enter general terms.
    Also I'd suggest you take a look at work being done by NASA related to
    space station impacts. (this goes back years)
    They have some video's and data on different materials
    and designs that may be of use that are freely available online.
    Some of the PDF's include the modeling so you could get some ideas there.

    PS. Also there is containment chambers for domestic aircraft work around.
    Last edited: Sep 20, 2013
  6. Baluncore

    Baluncore 2,525
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    Blast containment is normally applicable to explosions within enclosed spaces. Projectile penetration is probably best handled with a high tensile flexible fabric wall that wraps and holds the projectile.

    It will depend on;
    Whether the resulting deformation is elastic, (reusable), or plastic, (disposable).
    The support behind the sheet steel. Less support may be more. More time, less force.
    The nose profile and cross-section of the projectile.
    What the projectile is made from, a brittle or a plastic material.
  7. Q_Goest

    Q_Goest 2,966
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  8. AlephZero

    AlephZero 7,300
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    Or for a cheaper solution, just surround your 1.2mm steel box with sandbags (and don't forget to cover the top!)

    Personally I would do that anyway, even if my calculations said it wasn't necessary. One feature of things that go bang unexpectedly is that they haven't read any textbooks about how they are supposed to behave. (I speak from experience here!)
  9. It seems to me that you aren't looking for a method of containment so much as analytical validation of the containment shell failure.

    I certainly am no expert on projectile mechanics, but would this not be a standard projectile calculation (assuming there IS such a thing)? Using "worst case scenarios" would mean to me (1) vector of projectile motion is perpendicular to surface, (2) shape of projectile contacting the surface would be a sharp point for maximum stress of the material.

    You could probably drive yourself nuts working through respective coefficients of restitution to manage the respective deformations of projectile and containment material during collision. Or you could just take a first pass calculation and assume non-deformable projectile body. The stress alone from a pointed object would probably send that containment material way past the plastic zone anyway.

    This could be a very complex calculation, I think, because you'd have to account for a very dynamic situation:
    (1) determine the initial stress from impact
    (2) depending on shape of projectile, how does the kinetic energy of the projectile dissipate as it is turned into heat by the mechanical deformation of the containment material? ....which would change dynamically based on the changing cross-sectional impact profile of the projectile as it tries to pass through the containment material.

    Makes my head hurt just to think about an accurate analysis. A simplistic approach may yield a fast answer that negates all further analysis. Or you could go crazy with a finite element model of this scenario.

    I would probably approach this as a work-energy calculation, applying all kinetic energy to a point on the containment material and see what the max stress would be. If sufficiently high, then assume penetration.
  10. Bandit127

    Bandit127 205
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    I will try to put it another way.

    Your projectile has a kinetic energy of 3.5 kJ. That is apparently very close to the kinetic energy of a 7.92×57mm Mauser amour piercing rifle round.

    Assuming your projectile is steel then the analogy may hold.

    You may get better luck searching Google for bullet proof steel plate but a quick search for me mentions 1" (25 mm).
  11. Bobbywhy

    Bobbywhy 1,908
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    Just some anecdotal observation from Da Nang, Vietnam, Spring, 1968

    During the TET offensive against the invaders (The USA) many rockets rained down onto our airbase.
    After one 122 mm rocket attack I saw the result of the rocket warhead’s high velocity shrapnel: At the communal shower outside the enlisted men’s barracks, six hot water tanks in a row had been punctured by one small piece of shrapnel from the exploding rocket, and you could see daylight in a straight line through all six of the tanks.
  12. AlephZero

    AlephZero 7,300
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    If that is the case, we don't know most of the relevant information to answer the question. it depends very much on the shape of the projectile and the size and shape of the "room" - i.e. how likely is it to punch a hole through the steel plate, instead of absorbing energy by plastic deformation of the containment system and/or the projectile - not to mention large movements of the containment system transmitting energy into the surrounding air.
  13. This is all for a risk assessment (hence the worse case senerio) for a rotating disc bursting and a particle hitting the steel wall that it is enclosed in. I have assumed that the particle hits the wall perpendicular with a sharp edge.

    Thanks for all the feedback it has provided some useful reading and discussion. Looking at it all the Hagg and Sankey paper has provided the most information. I am looking for the most simple answer as possible on will it contain it and if not how thick the steel needs to be to contain it. Then I can start looking at possible other materials.

    Looking at the Hagg and Sankey equations you have E=0.5*M1*V^2*(1-(M1/(M1+M2))) so this calculates the energy that is absorbed by the plate when the projectile hits it. If this is more than the strain energy of the plate then it does penetrate it. I am right in the way I am understanding this equation??
  14. Baluncore

    Baluncore 2,525
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    One characteristic of a bursting disc, such as a grinding wheel, is that the pie shaped fragments are smaller than the original disc. The centrifugal force is shared by the fragments, which determines their trajectories. The rotational energy is shared by the fragments, which significantly increases their rate of rotation since they now have a reduced radius, (squared).

    The fragments are gyroscopically stabilised pie shaped cutters that, if they do not cut through the wall, will run around the inside of the room. They are best caught, rolled up and tangled in long fibres such as Kevlar. Chainsaw operators use fibre filled trousers that choke the chain if it cuts through the outer layer.

    Discs that fragment in an open area, convert their circular energy into linear velocity every time they touch the ground. It is not unexpected for parts of a green-waste shredder disc to travel for a km or more. The fragments will cut through motor vehicles and climb over more solid objects.

    The worst case for a disc is worse than you might at first expect.
  15. Q_Goest

    Q_Goest 2,966
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    It's been a while since I used this so please bear with me. I believe you are referencing equation 5 in that paper. That's the total energy after impact for the 2 parts; the projectile and the portion of the wall that's impacted as discussed in the paper. Some of the initial energy of the projectile is lost as is discussed in the paragraph immediately preceding this equation in the form of heat. Note that momentum is conserved which is how that lost energy is determined. This is "stage 1".

    Stage 2 is next. Once you determine the remaining energy after impact, you determine how much energy can be absorbed by the structure due to deformation of the body the projectile has impacted. So what you calculated determines the remaining energy that must be disipated due to strain and shear in the barrier.
  16. rollingstein

    rollingstein 438
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    Does 1.2 mm thick steel have strength sufficient to make a room out of it, forget projectile containment? I was surprised. Just hope it never corrodes a tiny bit even.
  17. Hello All,

    I am new here...

    I would like to know how to calculate the plastic flow stress in the Paper by SANKEY .

    What theorem to use for this will be of great help.

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