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

The answer is "583.042533 MPa".

Step-by-step explanation:

Solve the following for the real state strain 1:

`\varepsilon_(T)=\In (I_(il))/(I_(01))`

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

`K=(\sigma_(r1))/([\In (I_(il))/(I_(01))]^n)`

Solve the following for the true state stress and stress2.

`\sigma_(r2)=K(\varepsilon_(r2))^n`

`=(\sigma_(r1))/([\In (I_(il))/(I_(01))]^n) * [\In (I_(i2))/(I_(02))]^n\\\\=(399 \ MPa)/([In (54.4)/(47.7)]^(0.2)) * [In (57.8)/(47.7)]^(0.2)\\\\ =(399 \ MPa)/([ In (1.14046122)]^(0.2)) * [In (1.21174004)]^(0.2)\\\\ =(399 \ MPa)/([ In (1.02663509)]) * [In 1.03915873]\\\\=(399 \ MPa)/(0.0114161042) * 0.0166818905\\\\= 399 \ MPa * 1.46125948\\\\=583.042533\ \ MPa`