Question

To penetrate armor, a projectile's point concentrated force in a small area, creating a stress large enough that the armor fails. A species of jellyfish launches a pointed needle that can penetrate the hard shell of a crustacean. The rapid deceleration on impact creates a 32 μN force on the tip, which has a very small 15 nm radius.How does this compare to the ultimate stress of steel?

Answer 1

We know that the equiation of Stress is,

$\sigma = \frac{F}{A}$

Where

$F= Force$

$A= Area$

Here the Force is basically,

$F=32\mu N= 32*10^{-6}N$

And we know as well, that

$A= \pi r^2 = \pi 15*10^{-9}m$

So,

$\sigma = \frac{32.10^{-6}}{\pi (15*10^{-9})^2}$

$\sigma = 4.53*10^{10}N/m^2$

For this question, we know that the ultimate stress of steel is 1020Mpa

$\sigma_{steel}=1020Mpa=1020*10^6Pa$

So the ratio,

$R=\frac{\sigma}{\sigma_{steel}}=\frac{4.53*10^{10}}{0.084*10^{10}}$

$R= 44.38$