Steel ruptures when a shear of \(3 .5 \times 10^8\,\,N\,m^{-2}\) is applied. The force needed to punch a \(1\,cm\) diameter hole in a steel sheet \(0.3\,cm\) thick is nearly

the circum ference of the disk, and the total force exerted on its center only.

Letus assume that the shearing stress along the side surface of the disk is uniform, then

\(F > \int\limits_{Surface} {d{F_{\max }} = \int\limits_{surface} {{\sigma _{\max }}dA = {\sigma _{\max }}} \int\limits_{surface} {dA} } \)

\( = \int {{\sigma _{\max }}} .A = {\sigma _{\max }}.2\pi \left( {\frac{D}{2}} \right)h\)

\( = 3.5 \times {10^8} \times \left( {\frac{1}{2} \times {{10}^{ - 2}}} \right) \times 0.3 \times {10^{ - 2}} \times 2\pi \)

\( = 3.297 \times {10^4} = 3.3 \times {10^4}N\)