Calculating Stress Values for Cylindrical Steel Sample

[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.

Are my answers correct? Thanks.

 

[AlephZero]

The method looks OK.

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

 

[Chestermiller]

The method looks OK.

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

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