Steel ruptures when a shear of`3.5xx10^(8)Nm^(-2)` is applied. The force needed to punch a `1 cm` diameter hole in a steel sheet `0.3 cm` thick is nearly:

Assume that if the shear stress in steel exceeds about 4.00xx10^(8)N//m^(2) the steel ruptures. 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.

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.

Calculate the force F needed to punch a 1.46 cm diameter hole in a steel plate 1.27 cm thick. The ultimate shear strength of steel is 345 MN//m^2.

For steel Y=2xx10^(11) Nm^(-2) . The force required to double the length of a steel wire of area 1 cm^(2) is

Calculate the force F needed to punch at 1.46 cm diameter hole in a steel plate 1.27 cm thick fig. The ultimates shear stress of steel is 3.45 xx 10^(8)Nm^(-2) .

Calculate the force .F. needed to punch a 1.46cm diameter hole in a steel plate 1.27 cm thick (as shown in fig). The ultimate shear strength of steel is 345 M N//m^(2)

The force required to punch a hole of diameter 2 mm, will be (If a shearing stress 4xx10^(8)N//m^(2) .)