Question 6.11: A steel sheet, which has an R-value of 1.75 in all direction......
A steel sheet, which has an R-value of 1.75 in all directions in the sheet, was stretched in biaxial tension with \sigma_{z} = 0. Strain measurements indicate that throughout the deformation \varepsilon_{y} = 0. Find the stress ratio, α = \sigma_{y}/ \sigma_{x}, that prevailed according to the Hill yield criterion (Equation (6.33)) and according to the nonquadratic yield criterion (Equation (6.40)) with a = 8.
(\sigma _{y}-\sigma _{z})^2+(\sigma _{z}-\sigma _{x})^2 +R(\sigma _{x}-\sigma _{y})^2 = (R+1)X^2, (6.33)
(\sigma _{y}-\sigma _{z})^a+(\sigma _{z}-\sigma _{x})^a +R(\sigma _{x}-\sigma _{y})^a = (R+1)X^a, (6.40)
For the Hill criterion Equation (6.36) gives \alpha =\left[(R+1)\rho +R\right]/\left[ (R+1)+R\rho \right] . With ρ = 0, α = R/(R + 1) = 1.75/2.75 = 0.636.
For the nonquadratic criterion, the flow rules (Equation (6.41))
d\varepsilon_{x}:d\varepsilon_{y}:d\varepsilon_{z} = \left[R(\sigma _{x}-\sigma _{y})^{a-1} + (\sigma _{x}-\sigma _{z})^{a-1}\right]: \left[(\sigma _{y}-\sigma _{z})^{a-1} +R(\sigma _{y}-\sigma _{x})^{a-1} \right]:\left[(\sigma _{z}-\sigma _{x})^{a-1} + (\sigma _{z}-\sigma _{y})^{a-1}\right]. (6.41)
with \sigma_{z} = 0 and ρ = 0 can be expressed as 0=d\varepsilon_{y}/d\varepsilon_{x}=\left[\alpha ^{a-1} +R(\alpha -1)^{a-1}\right]/\left[R(1-\alpha )^{a-1} +1\right]=\left[\alpha ^7 -1.75(1-\alpha )^7\right]/ \left[1.75(1-\alpha )^7 + 1\right] . Trial and error solution gives α = 0.520.