# Calculating yield stress?

I have a cylindrical steel sample tested in tension. It’s cross-sectional area is 500 mm

^{2}. 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 x10^{3} / 5 x10^{-4} = 4.1 x10^{8} Nm^{-2}

**Ultimate Tensile Strength**

Ultimate Tensile Stress = 258 x10^{3} / 5 x10^{-4} = 5.16 x10^{8} 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 10^{3} / 3 x10^{-4} = 6.7 x10^{8} Nm^{-2}.

Are my answers correct? Thanks.

http://ift.tt/VobUWz

## Leave a comment