Question

For some metal alloy, a true stress of 415 MPa produces a plastic true strain of 0.475. How much does a specimen of this material elongate when a true stress of 325 MPa is applied if the original length is 300 mm

Answer 1

Answer:

$58.63\ \text{mm}$

Explanation:

$\sigma_t$ = True stress = 415 MPa

$\varepsilon_t$ = True strain = 0.475

$l_i$ = Original length of the alloy = 300 mm

Let us assume n = Strain hardening component = 0.25

K = Constant

True stress is given by

$\sigma_t=K(\varepsilon_t)^n\\\Rightarrow K=\dfrac{\sigma_t}{(\varepsilon_t)^n}\\\Rightarrow K=\dfrac{415}{0.475^{0.25}}\\\Rightarrow K=499.89\approx 500\ \text{MPa}$

Now $\sigma_t=325\ \text{MPa}$

$325=500(\varepsilon_t)^{0.25}\\\Rightarrow (\dfrac{325}{500})^{\dfrac{1}{0.25}}=(\varepsilon_t)\\\Rightarrow \varepsilon_t=0.1785$

True strain is given by

$\varepsilon_t=\ln(\dfrac{l_f}{l_i})\\\Rightarrow e^{\varepsilon_t}=\dfrac{l_f}{l_i}\\\Rightarrow l_f=l_ie^{\varepsilon_t}\\\Rightarrow l_f=300e^{0.1785}\\\Rightarrow l_f=358.63\ \text{mm}$

Elongation of material is $l_f-l_i=358.63-300=58.63\ \text{mm}$.