$ {\mathcal l}_\mathrm i=20\;\mathrm{mm}\;\;\;\;\;\;\;{\mathcal l}_\mathrm f=10\;\mathrm{mm} $
$ \in=\frac{\triangle\mathcal l}{{\mathcal l}_\mathrm i}=\frac{10-20}{20}=-0.5 $
True strain $ \left({\mathrm e}_\mathrm r\right)=\ln\left(1+\in\right)=\ln\left(1-0.5\right)=0.69 $
Loading...