Question 6.9: The strains measured on the surface of a piece of sheet meta......
The strains measured on the surface of a piece of sheet metal after deformation are \varepsilon_{1} = 0.182, \varepsilon_{2} = −0.035. The stress–strain curve in tension can be approximated by \sigma = 30 + 40\varepsilon . Assume the von Mises criterion and assume that the loading was such that the ratio of \varepsilon_{2}/\varepsilon_{1} was constant. Calculate the levels of \sigma _{1} and \sigma _{2} reached before unloading.
First, find the effective strain. With constant volume, \varepsilon_{3}=- \varepsilon_{1}-\varepsilon_{2}= − 0.182+0.035= −0.147. TheMises effective strain, \overline{\varepsilon }=\left[(2/3)(\varepsilon_{1}^{ \ \ 2}+\varepsilon_{2}^{ \ \ 2}+\varepsilon_{3}^{ \ \ 2})\right]^{1/2} =\left[(2/3) (0.182^2 +0.035^2 +0.147^2)\right]^{1/2} =0.193. [Note that this is larger than 0.182 and smaller than 1.15 × 0.182.] Since the tensile stress–strain curve is the effective stress–effective strain relation, \overline{\sigma }=30+40\overline{\varepsilon }=30+40\times 0.193 =37.7 . At the surface, \sigma _{z}=0 , so the effective stress function can be written as \overline{\sigma }/\sigma_{1} =(1/\sqrt{2} )\left[\alpha ^2 +1 +(1-\alpha )^2\right]^{1/2} = (1-\alpha +\alpha ^2)^{1/2}. From the flow rules with \sigma _{z}=0, \rho =d\varepsilon_{2}/ d\varepsilon_{1} =(\alpha -1/2)/(1-\alpha /2) . Solving for \alpha , \alpha =(\rho +1/2)/(1+\rho /2). Substituting \rho =-0.035/0.182=-0.192, \alpha =(-0.192+1/2)/(1- 0.192/2)=+0.341, \overline{\sigma }/\sigma_{1} =\left[1-\alpha +\alpha ^2\right]^{1/2} =\left[1-0.341+0.341^2\right]^{1/2} =0.881 . \sigma_{1}=0.881\sqrt{\sigma } =37.7/0.881 =42.8 and \sigma_{2} =\alpha \sigma_{1} =0.341 \times 42.8=14.6.