9. A specimen originally 0.5 inches in diameter is deformed until the diameter is 0.3 inches under an applied load of 5,000 lb. Find the true stress and the engineering stress. Solution: Engineering Stress = F/original area = 5,000 lb/ ( 0.52/4)in = 25,500 psi True stress =F/actual area = 5,000 lb/( 0.32/4)in = 70,700 psi 10. A specimen originally 10.0 cm in length is deformed by applying a tensile load. Compare engineering strain to true strain for deformations of 0.1 mm, 1.0 cm, and 10 cm. Solution: engineering strain 8L/L, true strain = In (LoL), and Lo = 8L + L Deformation True Strain Engineering Strain 0.1 mm 1.0 cm 0.0010 0.0953 0.0010 0.100 10 cm 0.693 2.0