Q. Steel ruptures when a shear of $ 3.5 \times 10^8 \,Nm^{-2}$ is applied. The force needed to punch a $1\, cm$ diameter hole in a steel sheet $0.3\, cm$ thick is nearly :
Solution:
$ \frac{ F }{ A } =$ stress $=3.5 \times 10^{8} N / m ^{2} $
$A =2 \pi rt $
$=2 \pi \times \frac{1}{2} \times 0.3 $
$=0.3 \pi \times 10^{-4} m ^{2} $
$ F = A \times$ stress
$=0.3 \pi \times 10^{-4} \times 3.5 \times 10^{8} $
$=1.05 \pi \times 10^{4}$
$=3.3 \times 10^{4} N $
