So, I am trying to calculate the linear strain of a shaft which is under plastic deformation in yielding region.
The diameter is changing from 2d to 1.5d
I'm assuming that the volumetric strain is zero.$(\epsilon_v=0)$ or in other words $\mu = 0.5$
Now, I've tried the following two methods and I'm getting two different answers:
Method 1 $$\epsilon_v = 0 \\ \Rightarrow \epsilon_l + 2\epsilon_d = 0 \\ \Rightarrow \epsilon_l = -2(\frac{1.5d - 2d}{2d}) \\ \Rightarrow \epsilon_l = 0.5$$ Method 2 $$V_i = V_f\\ \Rightarrow \frac{\pi}{4}d_i^2l_i = \frac{\pi}{4}d_f^2l_f\\ \Rightarrow \epsilon_l = \frac{\Delta l}{l_i} = \frac{7}{9}$$ Where $\epsilon_l$ is longitudinal strain, $\epsilon_d$ is lateral strain($\bot$ to longitudinal direction), subscripts $i, f$ refer to initial and final conditions respectively.
All the strains are engineering strains only.
Now ideally, both the methods should give same solution.
Where am I going wrong?