A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.20 is produced when a true stress of 575 MPa (83,500 psi) is applied; for the same metal, the value of K in Equation 7.19 is 860 MPa (125,000 psi). Calculate the true strain that results from the application of a true stress of 600 MPa (87,000 psi).

∈ₙ= 0.20      σₙ = 575 MPa

K = 860 MPa

the true strain that results from the application of a true stress of 600 MPa (87,000 psi) IS ∈ₙ  = 0.24

Explanation:

Given that ,∈ₙ= 0.20      σₙ = 575 MPa

K = 860 MPa

When  σₙ= 575 MPa, what is ∈ₙ

Use ,  σₙ= K ∈ˣₙ   , solve for X

575 MPa = 860 MPa(0.20)ˣ

Take log of each side

log(575) = log860 - X log(0.20)

X = 0.25

Plug in X :

600 MPa = 860 MPa(∈ₙ )

X=0.25

∈ₙ =

∈ₙ  = 0.24

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explanation: