Assume that if the shear stress in steel exceeds about 4.00 108Nm2, 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 1.05-cm-diameter hole in a steel plate 0.870 cm thick.

Question

Assume that if the shear stress in steel exceeds about 4.00 108Nm2, 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 1.05-cm-diameter hole in a steel plate 0.870 cm thick.

Jim Hunt · Accepted Answer

Step 1  Given,  Shearing stress, S=4×108Nm2  We know,  Shearing stress, S=Shearing force(F)Area(A)  So,  Shearing force, F=S×A  Step 2  a) Given,  diameter, d=1.45cm  radius, r=d2=0.725cm=0.72×10−2m  Thus,  Shearing force,  F=S×A  ⇒F=4×108×π×r2  ⇒D=4×108×π×(0.72×10−2)2  ⇒F=6.5×104N  Step 3  b) Given,  Diameter, d=1.05cm  Radius, r=0.525cm=0.52×10−2m  Thickness, t=0.870cm=0.87×10−2m  Thus,  Shearing force   F=S×A  ⇒F=4×108×2πr  ⇒F=4×108×2π×0.52×10−2×0.87×10−2  ⇒F=1.13×105N  Step 4  Answer:   a) Shearing Force =6.5×104N   b) Shearing Force =1.13×105N

Juan Spiller · Accepted Answer

Step 1  σ=4×108Nm2  a) d=1.45×10−2m  ∴γ=0.72×10−2m  ∴F=γA=4×108×π(0.72×10−2)2  ⇒F=π(4×108)(0.72×10−2)2=6.5×10+4N  b) d=1.05cm  ∴γ=0.52×10−2m  f=0.4×10−2m  ∴F=(4×108)(πγell)=4×108×(π×0.52×0.4)×10−4  ⇒F=2.61×104N

Assume that if the shear stress in steel exceeds about

Answered question

Answer & Explanation

New Questions in Other