Aparticular metal has a flow curve with parameters: strength coefficient = 35,000 psi and strain-hardening exponent = 0.26. a tensile specimen of the metal with gage length = 2.0 in is stretched to a length = 3.3 in. determine the flow stress at this new length and the average flow stress that the metal has been subjected to during deformation.

Flow stress (Yf):  Yf = 29,240 PSI

Average flow stress (Yf): Yf = 23206 PSI

Explanation:

K = 35000 PSI

L₀ = 2.0 in

L₁ = 3.3 in

n = 0.26

We can apply the following equations:

Engineering strain or Maximum strain (ε):

ε = ln (L₁ / L₀)

ε = ln (3.3 / 2.0) = ln (1.65) = 0.501

Flow stress (Yf):

Yf = K*εⁿ

Yf = (35000 PSI) (0.501)⁰·²⁶

Yf = 29,240 PSI

Average flow stress (Yf)

Yf = K*εⁿ / (1 + n)

Yf = (35000)(0.501)⁰·²⁶ / (1.26) = 23206 PSI

answer:

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explanation:

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answer:

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explanation: