Question:penetration of penetrator (projectile)

[yoque]

Dear All,

I have a problem and would like to solve it as soon as possible. Please find it below:

There is a long rod projectile and a semi-infinite target (1-D solution is fine)

Projectile velocity 4000m/s
Projectile density 7,5g/cm3
Long rod has a length of 50mm and diameter 5mm
Projectile yield strength 100MPa

Target density 8g/cm3
Target yield strength 500MPa

Calculate the penetration?

Could you please help me to solve this problem.

Thanks in advance for your efforts.

 

[Achy47]

Thank you for reaching out with your problem. I understand your urgency and I am happy to assist you in finding a solution.

To calculate the penetration of the long rod projectile into the semi-infinite target, we will need to use the projectile's kinetic energy formula: KE = 1/2 * m * v^2. In this case, the mass (m) of the projectile can be calculated by multiplying its density (7.5g/cm3) by its volume, which can be calculated using the formula for the volume of a cylinder: V = π * r^2 * h. In this case, the radius (r) of the long rod is 2.5mm and the height (h) is 50mm. Therefore, the mass of the projectile is 0.00294kg.

Next, we need to calculate the kinetic energy of the projectile using its velocity (4000m/s). This gives us a value of 23.52MJ.

To determine the penetration depth, we will use the formula for the depth of penetration: d = (m * v)/(σ * ρ), where m is the mass of the projectile, v is the velocity, σ is the yield strength of the target, and ρ is the density of the target.

Substituting the values given in the problem, we get: d = (0.00294kg * 4000m/s)/(500MPa * 8g/cm3) = 0.00000147m or 1.47 micrometers.

Therefore, the penetration depth of the long rod projectile into the semi-infinite target is 1.47 micrometers.

I hope this helps you solve your problem. If you have any further questions or need clarification, please do not hesitate to ask.