Question 6.10: Derive the flow rules for the high-exponent yield criterion,......
Derive the flow rules for the high-exponent yield criterion, Equation (6.29).
\left|\sigma_{2}-\sigma_{3}\right|^a + \left|\sigma_{3}-\sigma_{1}\right|^a +\left|\sigma_{1}-\sigma_{2}\right|^a =2Y^a. (6.29)
When the yield function is expressed as f=(\sigma_{y}-\sigma_{z})^a + (\sigma_{z}-\sigma_{x})^a +(\sigma_{x}-\sigma_{y})^a , the flow rules can be found from d\varepsilon_{ij} = d\lambda (df/d\sigma_{ij} ):d\varepsilon_{x}= d\lambda \left[a(\sigma _{x}-\sigma _{z})^{a-1} + a(\sigma _{x}-\sigma _{y})^{a-1}\right], \ d\varepsilon _{y}=d\lambda \left[a(\sigma _{y}-\sigma _{x})^{a-1}+a(\sigma _{y} \ \sigma _{z})^{a \ 1}\right], and d\varepsilon_{z} =d\lambda \left[a(\sigma _{z}-\sigma _{x})^{a-1} + a(\sigma _{z}-\sigma _{y})^{a-1}\right].
These can be expressed as
d\varepsilon_{x}:d\varepsilon_{y}:d\varepsilon_{z} = \left[(\sigma _{x}-\sigma _{z})^{a-1} + (\sigma _{x}-\sigma _{y})^{a-1}\right]: \left[(\sigma _{y}-\sigma _{x})^{a-1} +(\sigma _{y}-\sigma _{z})^{a-1} \right]:\left[(\sigma _{z}-\sigma _{x})^{a-1} + (\sigma _{z}-\sigma _{y})^{a-1}\right]. (6.31)