Tutorial Exercise Assume that if the shear stress in steel exceeds about 4.00 108 N/m, the steel ruptures. (a) Determine the shearing force necessary to shear a steel bolt 1.45 cm in diameter. (b) Determine the shearing force necessary to punch a 1.35-cm-diameter hole in a steel plate 0.900 cm thick. Step 1 (a) We know that force F is equal to the surface area A times stress so we have F = A = 4 4 108 N/m (0.725 = 6.6052 6.61 104 N. 0.725 x 10-2 m) Step 2 (b) The area over which the shear occurs is equal to the circumference of the hole d times its thickness h. Thus, F = A = (h)(d) 4 108 N/m) (0.900 102 m) +) ( 0.0068 = Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. x 10-2 m) 7.69061 = Your response differs from the correct answer by more than 10%. Double check your calculations. x 104 N. A=2rh