Elasticity — Question 10

Question 10

Unloading reveals permanent strain and unrecovered work

An ideal material loads linearly with $E=200\,\mathrm{GPa}$ to yield stress $200\,\mathrm{MPa}$, then deforms at constant stress to total strain $.004$. It unloads linearly with slope $E$ to zero stress. Use work per initial volume $w=\int\sigma\,d\epsilon$; neglect area changes.

1. Find yield strain and residual strain after unloading.

2. Find loading work, returned elastic work and unrecovered work per initial volume.

Question 10 - Solution

Unloading reveals permanent strain and unrecovered work

(a) Derivation and calculation

Yield occurs at $$\boxed{\epsilon_y=\sigma_y/E=(200\times10^6)/(200\times10^9)=.00100}.$$ Unloading recovers elastic strain $\sigma_y/E=.00100$, so $$\boxed{\epsilon_{\mathrm{residual}}=.00400-.00100=.00300}.$$

(b) Result and interpretation

Loading work comprises an elastic triangle and a plastic rectangle: $$w_{\mathrm{load}}=\tfrac12\sigma_y\epsilon_y+\sigma_y(.004-.001) =100+600=\boxed{700\ \mathrm{kJ/m^3}}.$$ Returned elastic work is $$w_{\mathrm{returned}}=\frac{\sigma_y^2}{2E}=\boxed{100\ \mathrm{kJ/m^3}}.$$ Therefore $$\boxed{w_{\mathrm{unrecovered}}=700-100=600\ \mathrm{kJ/m^3}}.$$

Physical check

The unrecovered work equals the yield stress times the permanent strain in this ideal perfectly plastic model.