Question

Assume that if the shear stress in steel exceeds about 4.00 $\times {10}^{8}N/m^2,$ the steel reptures. Determine. the shearing force necessary to (a) shear a steel bolt 1.00 cm in diameter and (b) punch a 1.00-cm-diameter hole in a steel plate 0.500 cm thick.

Answer 1

a. Force applied F = (A) (stress)
$\pi$(5.00 $\times {10}^{-3}m{)}^2(4.00\times {10}^{8}N/m^2)=3.14\times {10}^{4}N$
b. The area over which the shear stress occurs is equal to the circumference of the hole times its thickness.
Thus, A = (2 $\pi$ r) t = 2$\pi$ (5.00 $\times {10}^{-3}$ m) (5.00 $\times {10}^{-3}m)$
= 1.57 $\times {10}^{-4}{m}^2$
So, F = (A) (stress)
= (1.57 $\times {10}^{-4}{m}^2$) (4.00 $\times {10}^{8}N/m^2$) = 6.28 $\times {10}^{4}N$