A model for projectile penetration of thin shells

[baxtrom]

Hi there fellow desktop scientists and others

that topic some time ago when a game developer asked about realistic damage formulae inspired me to come up with an equation for projectile penetration of a thin elastic-plastic shell. In the derivation below I neglect losses due to deformation of the projectile, and further simply assume that the projectile cuts through the plate by means of shear stress (i.e. wadcutter vs thin shells). Any comments or suggestions?

 

I use the following symbols:

 

velocity [math]v[/math]

time [math]t[/math]

initial velocity [math]v_0[/math]

deceleration [math]c[/math]

shell thickness [math]T[/math]

force of resistance [math]F[/math]

projectile mass [math]m[/math]

initial kinetic energy of projectile [math]K[/math]

projectile diameter [math]d[/math]

shear stress [math]\tau[/math]

shell yield strength [math]S_y[/math]

 

Assume a linear relationship between velocity and deceleration, i.e. constant force of resistance, such that the exit velocity of the projectile is zero. Then,

 

[math]v(t) = v_0 - c t[/math],

 

where [math]c = \frac{v_0^2}{2 T}[/math] (this implies that [math]v = 0[/math] after the projectile has decelerated through a distance [math]T[/math]).The deceleration of the projectile is [math]c[/math] which results in a resistance force equal to

 

[math]F = m c = \frac{m v_0^2}{2 T} = \frac{K}{T}[/math].

 

Assume further that the projectile cuts a hole in the target shell by shear stress along the circumference of the projectile,

 

[math]\tau = \frac{F}{\pi d T}=\frac{K}{\pi d T^2}[/math].

 

According to Tresca theory (as preferred in the US!) yield will start when [math]2 \tau[/math] equals the yield strength of the shell material. Thus, as a condition for penetration we have

 

[math]\frac{2 K}{\pi d T^2} = S_y[/math].

 

The maximum shell thickness that the projectile penetrates becomes

 

[math]T = \sqrt{\frac{2 K}{\pi d S_y}}[/math].

 

This would imply that a 10 mm wadcutter, with a mass of 0.01 kg (10 gram) and traveling at 300 m/s (kinetic energy equal to 450 J), would be capable of penetrating approx. 9 mm of structural steel (350 MPa). Is that a realistic result?

Edited June 22, 2011 by baxtrom

[InigoMontoya]

Hi there fellow desktop scientists and others

that topic some time ago when a game developer asked about realistic damage formulae inspired me to come up with an equation for projectile penetration of a thin elastic-plastic shell. In the derivation below I neglect losses due to deformation of the projectile, and further simply assume that the projectile cuts through the plate by means of shear stress (i.e. wadcutter vs thin shells). Any comments or suggestions?

 

For hard stuff, your equation looks reasonable, but damned few bullet/target combinations are hard. In other words, you mention "elastic-plastic shell" but your equation completely neglects elastic deformation of the shell - which in the case of body armor, is the primary mode of energy absorbtion.

 

As an aside, you may want to talk to somebody at Survice Engineering (http://www.survice.com). It's been a long time, but they used to put out a little tool called "Barrier" that did nothing but the kinds of calculations you're concerning yourself with.

Edited June 23, 2011 by InigoMontoya

[baxtrom]

For hard stuff, your equation looks reasonable, but damned few bullet/target combinations are hard. In other words, you mention "elastic-plastic shell" but your equation completely neglects elastic deformation of the shell - which in the case of body armor, is the primary mode of energy absorbtion.

 

Come to think of it, steel also undergoes deformation hardening up until the tensile limit which typically is perhaps 50-100% higher than the yield strength for structural steels. Also as you point out, the equation does not consider elastic deformation. I imagine in reality a projectile would cause concentric waves propagating away from the area of impact, transporting energy and decreasing the penetration of the projectile. Also, the deformation of the projectile is completely neglected by my formula. However, the end result looks reasonable in one sense, since penetration increases with kinetic energy and decreases with caliber and shell strength. I guess the formula, if it has any correlation with reality at all, would be correct in the high velocity and thin shell limit, where I imagine shearing is the main mechanism of penetration.

 

Thanks for the link. Looks like the software is not available for direct download, and I don't think I'll bother them by asking for it.

Edited June 23, 2011 by baxtrom

[John Cuthber]

Sorry, but if I have understood that properly, I can easily show that it's incomplete at best.

It says that the thickness of armour I can get through increases as the strength of the shell falls. T is proportional to 1/ root Sy

 

That doesn't make much sense.

It means that if I had two shells, one made from steel and the other made from powdered lead, held together with glue so they had the same density but one was much stronger than the other, the weak one would penetrate better.

 

Water has practically no strength (i.e Sy = zero) so rain should penetrate an indefinitely thick layer of armour.

[baxtrom]

It says that the thickness of armour I can get through increases as the strength of the shell falls.

 

Sorry, actually you misunderstood but it's my fault Without thinking about it I used the term "shell" not for the projectile, but for the plate which is impacted. That was actually a bit stupid, but I'm used to the term shell in the context of "plate" (i.e. the target). So, the penetration decreases with target strength. The strength of the projectile is not considered in the formulas (assumed rigid if you like).

 

Thanks for pointing it out to me. I'll edit the post if I can.

Edit: I can't edit the post!

Edited June 23, 2011 by baxtrom

[InigoMontoya]

I imagine in reality a projectile would cause concentric waves propagating away from the area of impact, transporting energy and decreasing the penetration of the projectile.

In fibrous body armor, you can think of it as a concentric ring where the fibers have been pulled tight, are therefore stretching, and therefore absorbing energy. And since you brought such things up, be advised that the speed of sound within the fibers themselves is an important property when choosing an armor fiber as it drives how quickly that ring propagates. The farther it propagates, the more fibers are involved, the more energy absorbed....

Edited June 24, 2011 by InigoMontoya