I have a cylindrical steel sample tested in tension. It's cross-sectional area is 500 mm2. I am also told that: Force at yield = 205 kN Maximum force = 258 kN Force at fracture = 200 kN I am asked to calculate the yield stress, ultimate tensile stress and fracture stress if the reduction of area at fracture is 40%. Working: Yield Stress: Yield Stress = 205 x103 / 5 x10-4 = 4.1 x108 Nm-2 Ultimate Tensile Strength Ultimate Tensile Stress = 258 x103 / 5 x10-4 = 5.16 x108 Nm-2 Fracture Stress For this question I reduced the 5 x10-4 by 40% to get 3 x10-4. I then did 200 x 103 / 3 x10-4 = 6.7 x108 Nm-2. Are my answers correct? Thanks.