Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 412 MPa (59760 psi) is applied if the original length is 480 mm (18.90 in.)? Assume a value of 0.22 for the strain-hardening exponent, n.

Answer 1

the elongation of the metal alloy is 21.998 mm

Step-by-step explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 /
`0.02^{0.22`

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

`l_i` =
`l_0e^{ET`

given that
`l_0` = 480 mm

we substitute

`l_i` =
`480mm` ×
`e^{0.04481`

`l_i` = 501.998 mm

Now we find the elongation;

Elongation =
`l_i - l_0`

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm