# Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 7

## Problem 1: Part 7: Flow rule

The ${\displaystyle J_{2}}$ theory of plasticity also states that the material satisfies the von Mises yield condition

${\displaystyle f({\boldsymbol {\sigma }},\alpha ,T):={\sqrt {\cfrac {3}{2}}}\lVert \mathbf {s} \rVert _{}-\sigma _{y}(\alpha ,T)}$

where ${\displaystyle \mathbf {s} }$ is the deviatoric part of the stress ${\displaystyle {\boldsymbol {\sigma }}}$. Derive an expression for ${\displaystyle \partial f/\partial {\boldsymbol {\sigma }}}$ in terms of the normal to the yield surface

${\displaystyle \mathbf {n} ={\cfrac {\mathbf {s} }{\lVert \mathbf {s} \rVert _{}}}~.}$

The von Mises yield function is

${\displaystyle f={\sqrt {\cfrac {3}{2}}}~{\sqrt {\mathbf {s} :\mathbf {s} }}-\sigma _{y}~.}$

We can alternatively write the yield function as

${\displaystyle f={\sqrt {\mathbf {s} :\mathbf {s} }}-{\sqrt {\cfrac {2}{3}}}~\sigma _{y}~.}$

in which case the following equations take a slightly different form.

Therefore,

{\displaystyle {\begin{aligned}{\frac {\partial f}{\partial {\boldsymbol {\sigma }}}}=f_{\boldsymbol {\sigma }}&={\sqrt {\cfrac {3}{2}}}~{\frac {\partial }{\partial {\boldsymbol {\sigma }}}}\left({\sqrt {\mathbf {s} :\mathbf {s} }}\right)\\&=\left({\sqrt {\cfrac {3}{2}}}\right)\left({\frac {1}{2}}\right)\left({\cfrac {1}{\sqrt {\mathbf {s} :\mathbf {s} }}}\right){\frac {\partial }{\partial {\boldsymbol {\sigma }}}}\left(\mathbf {s} :\mathbf {s} \right)\\&=\left({\sqrt {\cfrac {3}{2}}}\right)\left({\frac {1}{2}}\right)\left({\cfrac {1}{\lVert \mathbf {s} \rVert _{}}}\right){\frac {\partial }{\partial {\boldsymbol {\sigma }}}}\left(\mathbf {s} :\mathbf {s} \right)\end{aligned}}}

The deviatoric part of ${\displaystyle {\boldsymbol {\sigma }}}$ is

${\displaystyle \mathbf {s} ={\boldsymbol {\sigma }}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\boldsymbol {\mathit {1}}}\qquad ~{\text{or}}~\qquad s_{ij}=\sigma _{ij}-{\frac {1}{3}}~\sigma _{kk}~\delta _{ij}}$

Therefore,

{\displaystyle {\begin{aligned}\mathbf {s} :\mathbf {s} &=\left({\boldsymbol {\sigma }}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\boldsymbol {\mathit {1}}}\right):\left({\boldsymbol {\sigma }}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\boldsymbol {\mathit {1}}}\right)\\&={\boldsymbol {\sigma }}:{\boldsymbol {\sigma }}-{\cfrac {2}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\boldsymbol {\sigma }}:{\boldsymbol {\mathit {1}}}+{\cfrac {1}{9}}~\left({\text{tr}}({\boldsymbol {\sigma }})\right)^{2}~{\boldsymbol {\mathit {1}}}:{\boldsymbol {\mathit {1}}}\end{aligned}}}

Now,

${\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\mathit {1}}}=\sigma _{ij}\delta _{ij}=\sigma _{ii}={\text{tr}}({\boldsymbol {\sigma }})}$

and

${\displaystyle {\boldsymbol {\mathit {1}}}:{\boldsymbol {\mathit {1}}}=\delta _{ij}\delta _{ij}=\delta _{ii}=3~.}$

Therefore,

${\displaystyle \mathbf {s} :\mathbf {s} ={\boldsymbol {\sigma }}:{\boldsymbol {\sigma }}-{\cfrac {2}{3}}~\left({\text{tr}}({\boldsymbol {\sigma }})\right)^{2}+{\cfrac {1}{3}}~\left({\text{tr}}({\boldsymbol {\sigma }})\right)^{2}={\boldsymbol {\sigma }}:{\boldsymbol {\sigma }}-{\cfrac {1}{3}}~\left({\text{tr}}({\boldsymbol {\sigma }})\right)^{2}~.}$

The derivative with respect to ${\displaystyle {\boldsymbol {\sigma }}}$ is

${\displaystyle {\frac {\partial }{\partial {\boldsymbol {\sigma }}}}(\mathbf {s} :\mathbf {s} )={\frac {\partial }{\partial {\boldsymbol {\sigma }}}}({\boldsymbol {\sigma }}:{\boldsymbol {\sigma }})-{\cfrac {1}{3}}{\frac {\partial }{\partial {\boldsymbol {\sigma }}}}\left[\left({\text{tr}}({\boldsymbol {\sigma }})\right)^{2}\right]={\frac {\partial }{\partial {\boldsymbol {\sigma }}}}({\boldsymbol {\sigma }}:{\boldsymbol {\sigma }})-{\cfrac {2}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\frac {\partial }{\partial {\boldsymbol {\sigma }}}}\left[{\text{tr}}({\boldsymbol {\sigma }})\right]}$

Let us use index notation to find the derivatives. In index notation,

${\displaystyle {\frac {\partial }{\partial {\boldsymbol {\sigma }}}}({\boldsymbol {\sigma }}:{\boldsymbol {\sigma }})={\frac {\partial }{\partial \sigma _{ij}}}(\sigma _{kl}\sigma _{kl})={\frac {\partial }{\partial \sigma _{ij}}}\left(\sigma _{11}^{2}+\sigma _{12}^{2}+\sigma _{13}^{2}+\sigma _{21}^{2}+\sigma _{22}^{2}+\sigma _{23}^{2}+\sigma _{31}^{2}+\sigma _{32}^{2}+\sigma _{33}^{2}\right)}$

Hence, the components of the second-order tensor are

{\displaystyle {\begin{aligned}{\frac {\partial }{\partial \sigma _{11}}}(\sigma _{kl}\sigma _{kl})&=2~\sigma _{11}&&\qquad {\frac {\partial }{\partial \sigma _{12}}}(\sigma _{kl}\sigma _{kl})&=2~\sigma _{12}&&\qquad {\frac {\partial }{\partial \sigma _{13}}}(\sigma _{kl}\sigma _{kl})&=2~\sigma _{13}\\{\frac {\partial }{\partial \sigma _{21}}}(\sigma _{kl}\sigma _{kl})&=2~\sigma _{21}&&\qquad {\frac {\partial }{\partial \sigma _{22}}}(\sigma _{kl}\sigma _{kl})&=2~\sigma _{22}&&\qquad {\frac {\partial }{\partial \sigma _{23}}}(\sigma _{kl}\sigma _{kl})&=2~\sigma _{23}\\{\frac {\partial }{\partial \sigma _{31}}}(\sigma _{kl}\sigma _{kl})&=2~\sigma _{31}&&\qquad {\frac {\partial }{\partial \sigma _{32}}}(\sigma _{kl}\sigma _{kl})&=2~\sigma _{32}&&\qquad {\frac {\partial }{\partial \sigma _{33}}}(\sigma _{kl}\sigma _{kl})&=2~\sigma _{33}\end{aligned}}}

Therefore,

${\displaystyle {\frac {\partial }{\partial \sigma _{ij}}}(\sigma _{kl}\sigma _{kl})=2~\sigma _{ij}\qquad {\text{or}}\qquad {\frac {\partial }{\partial {\boldsymbol {\sigma }}}}({\boldsymbol {\sigma }}:{\boldsymbol {\sigma }})=2~{\boldsymbol {\sigma }}~.}$

Similarly,

${\displaystyle {\frac {\partial }{\partial {\boldsymbol {\sigma }}}}\left[{\text{tr}}({\boldsymbol {\sigma }})\right]={\frac {\partial }{\partial \sigma _{ij}}}(\sigma _{kk})={\frac {\partial }{\partial \sigma _{ij}}}(\sigma _{11}+\sigma _{22}+\sigma _{33})}$

Hence, the components of the second-order tensor are

{\displaystyle {\begin{aligned}{\frac {\partial }{\partial \sigma _{11}}}(\sigma _{kk})&=1&&\qquad {\frac {\partial }{\partial \sigma _{12}}}(\sigma _{kk})&=0&&\qquad {\frac {\partial }{\partial \sigma _{13}}}(\sigma _{kk})&=0\\{\frac {\partial }{\partial \sigma _{21}}}(\sigma _{kk})&=0&&\qquad {\frac {\partial }{\partial \sigma _{22}}}(\sigma _{kk})&=1&&\qquad {\frac {\partial }{\partial \sigma _{23}}}(\sigma _{kk})&=0\\{\frac {\partial }{\partial \sigma _{31}}}(\sigma _{kk})&=0&&\qquad {\frac {\partial }{\partial \sigma _{32}}}(\sigma _{kk})&=0&&\qquad {\frac {\partial }{\partial \sigma _{33}}}(\sigma _{kk})&=1\end{aligned}}}

Therefore,

${\displaystyle {\frac {\partial }{\partial \sigma _{ij}}}(\sigma _{kk})=\delta _{ij}\qquad {\text{or}}\qquad {\frac {\partial }{\partial {\boldsymbol {\sigma }}}}\left[{\text{tr}}({\boldsymbol {\sigma }})\right]={\boldsymbol {\mathit {1}}}~.}$

Plugging the above results into the expression for the derivative of ${\displaystyle \mathbf {s} :\mathbf {s} }$ we get

${\displaystyle {\frac {\partial }{\partial {\boldsymbol {\sigma }}}}(\mathbf {s} :\mathbf {s} )=2~{\boldsymbol {\sigma }}-{\cfrac {2}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\boldsymbol {\mathit {1}}}=2~\left({\boldsymbol {\sigma }}-{\cfrac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\boldsymbol {\mathit {1}}}\right)=2~\mathbf {s} ~.}$

Hence, we get

${\displaystyle {\frac {\partial f}{\partial {\boldsymbol {\sigma }}}}=f_{\boldsymbol {\sigma }}=\left({\sqrt {\cfrac {3}{2}}}\right)\left({\frac {1}{2}}\right)\left({\cfrac {1}{\lVert \mathbf {s} \rVert _{}}}\right)~2~\mathbf {s} ={\sqrt {\cfrac {3}{2}}}~{\cfrac {\mathbf {s} }{\lVert \mathbf {s} \rVert _{}}}={\sqrt {\cfrac {3}{2}}}~\mathbf {n} }$

The required expression is

${\displaystyle {\frac {\partial f}{\partial {\boldsymbol {\sigma }}}}=f_{\boldsymbol {\sigma }}={\sqrt {\cfrac {3}{2}}}~\mathbf {n} ~.}$