Assume that if the shear stress in steel exceeds about `4.00xx10^(8)N//m^(2)` the steel reptures. Determine the shearing force necessary to `a` sheal 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.

Explanation

a. Force applied `F=(A)` (stress)
`pi(5.00xx10^(-3)m)^(2) (4.00xx10^(8)N//m^(2))=3.14xx10^(4)N`

b. The area over which the shear stress ocurs is equal to the circumference of the hole times its thickness. Thus,
`A=(2pir)t=2 pi(5.00xx10^(-3)m)(5.00xx10^(-3)m)`
`=1.57xx10^(-4)m^(2)`
So, `F=(A)` (stress)`=(1.57xx10^(-4)m^(2))(4.00xx10^(8)N//m^(2))`
`=6.28xx10^(4)N`