Register to reply 
Projectile Blast Containment Calculations 
Share this thread: 
#1
Sep1913, 08:10 AM

P: 3

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
Sep1913, 09:27 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 6,755

1.2 mm blast containment ain't much containment. You might as well use a beer can.



#3
Sep2013, 02:34 AM

P: 3

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.



#4
Sep2013, 08:48 AM

P: 37

Projectile Blast Containment Calculations
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. 


#5
Sep2013, 09:09 AM

Sci Advisor
Thanks
P: 1,958

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 crosssection of the projectile. What the projectile is made from, a brittle or a plastic material. 


#6
Sep2013, 11:12 AM

Sci Advisor
HW Helper
PF Gold
P: 2,908

The reference I've used for burst containment of rotating machinery is:
Hagg, A. C., & Sankey, G. O. (1974). The containment of disk burst fragments by cylindrical shells. Journal of Engineering for Power, 96, 114. It can be found online here: http://gasturbinespower.asmedigitalc...icleid=1418505 You might also try Google Scholar and see what else comes up. http://scholar.google.com/scholar?hl...2C39&as_sdtp=# 


#7
Sep2013, 04:39 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 7,293

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!) 


#8
Sep2113, 07:08 AM

P: 159

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 nondeformable 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 crosssectional 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 workenergy 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. 


#9
Sep2113, 12:01 PM

P: 190

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). 


#10
Sep2113, 05:01 PM

PF Gold
P: 1,909

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. 


#11
Sep2113, 07:43 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 7,293




#12
Sep2413, 05:39 AM

P: 3

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?? 


#13
Sep2413, 08:34 AM

Sci Advisor
Thanks
P: 1,958

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 greenwaste 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. 


#14
Sep2413, 11:36 AM

Sci Advisor
HW Helper
PF Gold
P: 2,908

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. 


#15
Sep2813, 09:58 AM

PF Gold
P: 330

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. 


#16
Jan1014, 06:24 AM

P: 1

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. 


Register to reply 
Related Discussions  
Calculating Potential Energy of a Bow + Projectile calculations  Introductory Physics Homework  1  
Air Cannon Projectile Velocity Calculations...help please..  General Physics  1  
Containment problem  Set Theory, Logic, Probability, Statistics  5  
Projectile calculations including air resistance and weight  General Physics  2  
Projectile motion calculations from data gathered during trials  Introductory Physics Homework  0 