Find the Minimum diameter of a steel wire, which is used to raise a load of 4000 N if the stress is not exceed 95MN/sq(m)?

Detailed Solution:

Stress = 95 x 10^6 N/sq(m) = 95 N/sq(mm) = force/Area also Area = Force/Stress = 4000/95 =42.105 sq(mm) also Area = 0.785xdiameterxdiameter hence diameter = (42.105/0.7850^0.5 =(53.64)^0.5 =7.32 mm = 0.732 cm