I would like to ask you this question which I struggle to answer.

If I have a plate of some material X to which I am shooting at with a tank main gun with ammunition parameters: weight $= 10\text{kg}$, velocity $= 1550 \text{m/s}$, length $= 0.8\text{m}$, tip area $= 1\text{mm}^2$.

Now I calculate the pressure which this round produces at impact like this: $$P = \frac{F}{S} = \frac{10\times\frac{1550^2}{1.6}}{0.000001} = 1.5\times10^{13}.$$

So if I wanted this plate to have no plastic deformation (permanent deformation) after impact then this material must have yield strength of $1.5\times10^{13}$.
Am I right so far?

But if I would like to calculate this in terms of toughness that material X must have instead of yield strength, then I calculate the kinetic energy of this round which is $12 \text{MJ}$. Now how can I calculate over which area is this $12\text{MJ}$ distributed to get the final $\text{J/m}^3$ for required toughness? Since $1 \text{Pa} = 1 \text{J/m}^3$ then can it be said that the required toughness must also be at least $1.5\times10^{13}\text{J/m}^3$ Is that right or should all of this be calculated differently?

Thank you very much,

have a nice day.

  • 2
    $\begingroup$ I understand that this is a theoretical rather than a practical question, but regardless of the magic material it’s hitting, the tank round would not support that much tip pressure—it would flatten on impact, spreading the force over an area much larger than 1 mm^2. $\endgroup$ – Ben51 Dec 5 '18 at 23:57
  • $\begingroup$ Yes I knew that but there is no way of calculating the expansion so it is excluded from this calculation and therefore the result can be thought of as a high end result $\endgroup$ – Pan Mrož Dec 6 '18 at 0:22
  • 1
    $\begingroup$ How did you calculate the area of contact? This where you need to call upon the Hertzian contact theory to find the actual stress under a contact. $\endgroup$ – ja72 Dec 6 '18 at 1:10

To avoid permanent damage, you need to guard the surface against brineling. This means that there is a limit of contact pressure beyond which subsurface stresses exceed yield and the surface creates an indentation.

Let's say you know the brineling limit $P_{\rm max}$. Next you need to use Hertzian contact theory to calculate the force needed to reach this level. In the end you will get a pressure-force function of the form $$P \propto F^{1/3} $$ for a sphere on flat type of contact, as well as a displacement-force function of the form $$ \delta \propto F^{2/3} $$

But what you have is not a static case, but a projectile with certain kinetic energy. At peak deformation, the kinetic energy equals the elastic energy of the contact

$$ \frac{1}{2} m v^2 \propto \int \limits_0^\delta F\,{\rm d}x \propto \tfrac{2}{5} \delta ^ {5/2} $$

or the maximum dispacement being $\delta \propto m^{2/5} v^{4/5}$ and the force $F \propto m^{3/5} v^{6/5}$ .

Finally the contact pressure is $$ P \propto m^{1/5} v^{2/5}$$

So that is how you find how strong a plate is against brinelling.

  • $\begingroup$ I am sory, but what does that mean that P is proportional to m1/5v2/5? May I ask you how to calculate the exact pressure that the projectile exerct upon impact ? $\endgroup$ – Pan Mrož Dec 6 '18 at 17:00
  • $\begingroup$ So may I ask you is the yield strength of 1.5*10^13 Pa correct or must the plate have different strength? Thank you really much for an answer $\endgroup$ – Pan Mrož Dec 6 '18 at 17:05
  • $\begingroup$ Look at slide 7-8 in the attached pdf about Hertzian stress. You see that maximum von Mises stress is about 0.97 of contact pressure. If you know your yield strength then max pressure is 1/0.97 of yield. That is only for spherical contact. You might have other contact conditions. $\endgroup$ – ja72 Dec 6 '18 at 17:17
  • $\begingroup$ and is the yield strength calculation with result of 1.5*10^13 Pa correct? If so then this plate can withstand pressure of 0.97% of 1.5*10^13 right? $\endgroup$ – Pan Mrož Dec 6 '18 at 17:20
  • $\begingroup$ No, your calculation is not correct because it assumes fixed contact area. But the contact area changes with load. In addition, it assumes the pressure if constant throughout the contact, which is not either. $\endgroup$ – ja72 Dec 6 '18 at 18:14

Hi welcome to Physics SE.

I'm not sure I agree with the first part totally : the force exerted on the amunition is $F= \Delta p /\Delta t $. $\Delta p=m v$ and u're assuming $\Delta t= L / v$ which is a good scaling but not an exact result I think. But why do u put a factor 1/2 ?

Concerning the second reasoning, it is also an approximation, but it looks ok.

EDIT : Actually if we make the approximation that the velocity descreases linearly with respect to the time : $v(l)=v_0 (1- l/L)$. The mean velocity over $L$ is $v=v_0/2$. Then we can write $\Delta t = 2 L /v_0$.

  • $\begingroup$ well I calculate the force F = ma where a = v^2 / 2*d so that is (1550*1550)/(2*0.8) so deacceleration F = 10*1501562.5 = 15015625 N. Is that calculation right for the force that is exerted upon the plate? $\endgroup$ – Pan Mrož Dec 5 '18 at 23:31
  • $\begingroup$ And so is the calculation right that for the plate to sustain no pernament damage it must have yield strength of atl least 1.5*10^13 Pa? $\endgroup$ – Pan Mrož Dec 5 '18 at 23:35
  • $\begingroup$ Yeap but why are u writing the acceleration like that ? That's what I didn't really understand $\endgroup$ – J.A Dec 6 '18 at 8:09
  • $\begingroup$ I found that somewhere. :) But the important think is that the required strength is 1.5*10^13 Pa right? $\endgroup$ – Pan Mrož Dec 6 '18 at 17:09
  • $\begingroup$ I would be interested to have the reference. I would say it's the right value, but more because of the second reasoning. But I feel like everything here is only going to give you some approximation. So truly one can't say it's 1.5E13 or 2E13 or 1E13, just that it's something around that. But apparently u need the value for some homework or something (??)... $\endgroup$ – J.A Dec 6 '18 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.