Write the approach to solution also A tensile specimen with

Write the approach to solution also!

A tensile specimen with a 0.505 in initial diameter and 2.0 in gage length reaches maximum load at 20,000 lb and fractures at 16,000 lb. The minimum diameter at fracture is 0.425 in. a) Determine the engineering stress at maximum load and the true fracture stress, b) Determine the true strain at fracture, c) What is the engineering strain at fracture?

Solution

Solution a –Engineering stress at maximum load and True fracture stress

Engineering stress = P/A = 20000/ (3.14*0.505^2)/4 =99902.78 psi

True fracture stress = P/A = 16000/ (3.14*0.425^2)/4 =112842.43 psi

Solution b-True strain at fracture

True strain at fracture Ef = ln(Ao/Af) = ln( 0.505/.425) = 0.17246

Solution c - Engineering strain at fracture

Engineering strain at fracture

                    E= ln(1+e)

                  exp E = (1+e)

                 exp(0.17245) = 1+Ef

                 1.188224 -1=Ef

Engineering strain at fracture (Ef) =0.188224