Question

For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking. On the basis of this information, what engineering stress (in MPa) is necessary to produce an engineering plastic strain of 0.250?

Answer 1

The missing stresses in the question:

Engineering Engineering strain

Stress (MpA)

235 0.194

250 0.296

Answer:

σ = 245MPa

Explanation:

For this problem, we first need to convert engineering stresses and strains to true stresses and strains so that the constants K and n may be determined. Since σT = σ(1 + ε)

So,

σT1 = (235 MPa)(1 + 0.94) = 280 MPa

σT2 = (250 MPa)(1 + 0.296) = 324MPa

Similarly for strains, since;

εT = ln(1 + ε)

Thus;

ε T1 = ln (1 + 0.194) = 0.177

εT2 = ln (1 + 0.296) = 0.259

We know that true stress;

σε =(σT/K)^(1/n)

So taking log of both sides, to get;

log σT = log K + n log εT

which allows us to set up two simultaneous equations for the above pairs of true stresses and true strains, with K and

n as unknowns.

Thus ,

log (280) = log K + n log (0.177)

and

log (324) = log K + n log (0.259)

Solving both log equations simultaneously, we get;

K = 543 MPa and n = 0.383

Now converting, ε to true strain;

ε T= ln (1 + 0.25) = 0.223

From the true stress equation earlier quoted, we'll make σT the subject and get;

σT = (εT^n) x k

σT = 543 x (0.223^(0.383)) = 306 MPa

Now converting this value of σT

to an engineering stress using the first equation quoted in this answer, gives;

σ = σT/(1 +ε) = 306/(1+0.25) = 245MPa