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Sagot :
Answer:
the elongation of the metal alloy is 21.998 mm
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 / [tex]0.02^{0.22[/tex]
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
[tex]l_i[/tex] = [tex]l_0e^{ET[/tex]
given that [tex]l_0[/tex] = 480 mm
we substitute
[tex]l_i[/tex] =[tex]480mm[/tex] × [tex]e^{0.04481[/tex]
[tex]l_i[/tex] = 501.998 mm
Now we find the elongation;
Elongation = [tex]l_i - l_0[/tex]
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
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