Question

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 578 MPa (83830 psi) is applied; for the same metal, the value of K in the equation is 860 MPa (124700 psi).
Calculate the true strain that results from the application of a true stress of 600 MPa (87020 psi).

Answer 1

Answer:

The true strain that results from the application of a true stress of 600 MPa, will be 1.25.

Explanation:

The relation between true stress and true strain is given by the following formula:

σ = K (ε)^n

First, we find the value of n by substituting the values found by first test:

σ = 578 MPa

K = 860 MPa

ε = 0.2

Therefore,

578 MPa = (860 MPa)(0.2)^n

0.2^n = 0.67209

taking ln on both sides:

n ln(0.2) = ln(0.67209)

n = -0.3973/ln(0.2)

n = -1.6

Now, for the new stress value of 600 MPa, we calculate the strain:

600 MPa = (860 MPa)(ε)^-1.6

(ε)^1.6 = 1.4333

Taking power 1/1.6 on both sides, we get:

ε = 1.25

Answer 2

Answer:

0.234

Explanation:

True stress is ratio of instantaneous load acting on instantaneous cross-sectional area

σ = k × (ε)^n  

σ = true stress

ε = true strain

k = strength coefficient

n = strain hardening exponent

ε = ( σ / k) ^1/n

take log of both side

log ε = $\frac{1}{n}$ ( log σ  - log k)

n = ( log σ  - log k) / log ε

n = (log 578 - log 860) / log 0.20 = 0.247

the new ε = ( 600 / 860)^( 1 / 0.247) = 0.234