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2 years ago

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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 t

his 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.

1 answer:

Strike441 [17]2 years ago

6 0

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 / $0.02^{0.22$

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

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Explanation:

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To find the minimum temperature:

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3 years ago

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$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

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So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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