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?


1 Answer 1


I think that neither of your solutions is correct.

a) you are assuming that the volumetric strain is zero. However, if there is a change of volume then the volumetric strain can never be zero.

Also, in that scenario, I am a bit uncertain what is $\varepsilon_d$. Is it deviatoric strain or is it strain perpendicular to the longitudical direction and due to the symmetry you assume that its equal.

b) you assume that the volume is the same. However, due to the internal stresses and strains the volume might change (compressibility of material). It's like having a baloon in water (the lower you go i.e. the higher the hydrostatic stress, the volume of air reduces).

So neither I believe is correct.

  • $\begingroup$ I've assumed volumetric strain is zero because the material is under yielding(plastic deformation). $\epsilon_d$ is the strain perpendicular to longitudinal direction of the shaft(a cylindrical object) $\endgroup$
    – Anivarth
    Commented Sep 1, 2020 at 15:04
  • $\begingroup$ regarding $\epsilon_d$ a more appropriate sign would then be $\epsilon_r$ to denote strain in the radial direction. It is best if you try to solve this problem using cylindrical coordinates. $\endgroup$
    – NMech
    Commented Sep 1, 2020 at 16:55
  • $\begingroup$ The plastic component of the strain has no volume change, but the elastic component does have a volume change. $\endgroup$
    – alephzero
    Commented Sep 1, 2020 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.