1. (P3.2 Edition 2) A tensile specimen is machined to a gauge diameter of 0.357
ID: 1766704 • Letter: 1
Question
1. (P3.2 Edition 2) A tensile specimen is machined to a gauge diameter of 0.357 inch, and marked with a gauge length of 2.0 inch. When subject to a tension test, the following results were obtained Yield Load 2,000 lbf Diameter at ultimate load = 0.310 inch After completing the test, you are informed that the tensile specimen had been cold-worked a small amount before it was machined and tested, and in the annealed state, the flow curve of material could be described by = K(e)n with n = 0.5. a. b. c. What is the yield strength of the material? How much strain was induced by the unknown amount of cold work? What is the maximum load reached during the test?Explanation / Answer
Yield Strength =K€^-n
Assuming k as Expression =700
Then , yield Strength =2000 Ibf.
K=yield strength/(Strain)^n
K=(2000/0.693^0.5)
K =2403.8
Than,
Su=(Kn^n)(Au/Ao)
Au=(0.785*0.310^2)
Ao=(0.785*.357^2*2)
=(2403.8*0.5^0.5*(.310)^2/(.357^2*2)
Su=641.5 Mpa
SuAo=Stress at ultimate load * Au
Stress at ultimate Load= 641*(0.357)^2*2/(0.310^2*2)
= 1701.53 Mpa.
2.) Strain = (Stress at Ultimate load/K)^0.5
= (1701.53/2403.8)^0.5
=0.8408
Than The unknown Strain must be (Strain –Original Strain)= (0.8408-.693)
=0.1478
Which correspond to cold Work.
3. ) The Maximum Load can be Expressed as
Stress at ultimate Load =K(Strain)^n.
=2403.8(0.8408)^0.5
= 22040.04 Mpa.