Question

Check Your Understanding: True Stress and Stress A cylindrical specimen of a metal alloy 47.7 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 399 MPa causes the specimen to plastically elongate to a length of 54.4 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 47.7 mm to a length of 57.8 mm.

Answer 1

Answer:

The answer is "583.042533 MPa".

Explanation:

Solve the following for the real state strain 1:

$\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}$

Solve the following for the real stress and pressure for the stable.$\sigma_{r1}=K(\varepsilon_{r1})^{n}$

$K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}$

Solve the following for the true state stress and stress2.

$\sigma_{r2}=K(\varepsilon_{r2})^n$

     $=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa$