Question

A Cu-30% Zn alloy tensile bar has a strain hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm and an initial gage length of 3 cm, fails at an engineering stress of 120 MPa. At the moment of fracture, the gage length is 3.5 cm and the diameter is 0.926 cm. No necking occurred. Calculate the true stress when the true strain is 0.05 cm/cm.

Answer 1

79.74*10^6 Pa

Step-by-step explanation:

Based on the parameters provided, we have:

ε
`_(t)` = ln(
`l_(f)/l_(i)`)

Where initial gauge length = 3 cm and the final gauge length is 3.5 cm. Therefore:

ε
`_(t)` = ln(3.5/3) = ln(1.167) = 0.154

Similarly,

σ
`_(E)` = F/[3.142*(di^2)/4]

Where σ
`_(E)` = 120*10^6 Pa and di = 1 cm = 0.01 m

Therefore,

F = 120*10^6 * 3.142*(0.01^2)/4 = 9426 N

σ
`_(t)` = F/[3.142*(df^2)/4 = 9426/[3.142*(0.00926^2)/4 = 9426/6.74*10^-5 = 139.95*10^6 Pa

σ
`_(t)` = k*ε
`_(t) ^(0.5)` = 139.95*10^6

k = 139.95*10^6/(0.154)^0.5 = 356.63*10^6 Pa

Therefore, when ε
`_(t)` = 0.05 cm/cm

σ
`_(t)` = 356.63*10^6 (0.05)^0.5 = 79.74*10^6 Pa