In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of 90.8 N is applied perpendicular to each end. If the diameter of the wire is 1.84 mm, what is the breaking stress of the alloy?

Solution 39E Mathematically, the relationship between tensile force and stress is given by Stress = Tensile force /Area…..(1) Given, diameter of the wire = 1.84 mm Its radius = 0.92 mm = 0.92 × 10 3m 3 2 2 Area of cross-section of the wire A = × (0.92 × 10 ) m 6 2 A = 2.65 × 10 m Tensile force F = 90.8 N Therefore, from equation (1), Stress = 90.8 N/(2.65 × 106m )2 7 Stress = 3.41 × 10 Pa 7 Therefore, the breaking stress of the alloy is 3.41 × 10 Pa.