Question

If the shear stress in steel exceeds $4.00\times {10}^{8}N/m^2$ , the steel ruptures. The shearing force required to punch a $1cm$ diameter hole in a steel plate $0.5cm$ thick will be

• A. $3.28\times {10}^{4}N$
• B. $4.28\times {10}^{4}N$
• C. $5.28\times {10}^{4}N$
• D. $6.28\times {10}^{4}N$

Answer 1

The correct option is D

$6.28\times {10}^{4}N$

We have,

Force = (Stress)$\times$(area undergoing shear)

$F=Stress\times (\pi d\times t)$

On substitution,

$F=4\times {10}^8\times 3.14\times {10}^{-2}\times 0.5\times {10}^{-2}$

$F=6.28\times {10}^{4}N$