Question

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 :

• A. $1.4 \times 10^4\, N$
• B. $2.7 \times 10^4\, N$
• C. $3.3 \times 10^4\, N$
• D. $1.1 \times 10^4\, N$

Answer 1

Answer: (C)

$3.3 \times 10^4\, N$

Answer 2

$\frac{ F }{ A } =$ 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 $=0.3 \pi \times 10^{-4} \times 3.5 \times 10^{8}$ $=1.05 \pi \times 10^{4}$ $=3.3 \times 10^{4} N$