Steel ruptures when a shear of 3.5 × 10^8 ~N ~m^ -2 is applied. The...

Shearing strain is created along the side surface of the punched disk. Note that the forces exerted on the disk are exerted along the circumference of the disk, and the total force exerted on its center only. Let us assume that the shearing stress along the side surface of the disk is uniform, then $$\begin{aligned}\mathrm{F}>& \int_{\text{surface }} \mathrm{dF}_{\max }=\int_{\text{surface }} \sigma_{\max } \mathrm{dA}=\sigma_{\max } \int_{\text{surface }} \mathrm{dA}\\ &=\int \sigma_{\max } \cdot \mathrm{A}=\sigma_{\max } \cdot2 \pi\left(\frac{\mathrm{D}}{2}\right) \mathrm{h}\\ &=3.5 \times10^8 \times \ \left(\frac{1}{2} \times10^{-2}\right)\times0.3 \times10^{-2} \times2 \pi\\ &=3.297 \times10^4 \simeq 3.3 \times10^4 \mathrm{ N} \end{aligned}$$