# Calculating yield stress?

1. Jul 4, 2014

### Dave_

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.

2. Jul 4, 2014

### AlephZero

The method looks OK.

Stress values are usually given in MPa, not Nm-2.

3. Jul 4, 2014

### Staff: Mentor

Hi AlephZero,

Isn't fracture stress usually reported based on the original cross section (so called engineering stress), rather than the actual cross section (so-called true stress)? That would be consistent with how yield stress and ultimate stress were calculated by the OP. What is your experience in this regard?

Chet