The
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f11f387ee2af1216e2ea73fa79092836d4b81815)
where
is the deviatoric part of the stress
. Derive an expression for
in terms of the normal to the yield surface
![{\displaystyle \mathbf {n} ={\cfrac {\mathbf {s} }{\lVert \mathbf {s} \rVert _{}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2226abfc27664cf0477aae6060e321f3a5f63c26)
The von Mises yield function is
![{\displaystyle f={\sqrt {\cfrac {3}{2}}}~{\sqrt {\mathbf {s} :\mathbf {s} }}-\sigma _{y}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fb6c44270168a86fd7c6e3a33eb9df7c3d6cf60)
We can alternatively write the yield function as
![{\displaystyle f={\sqrt {\mathbf {s} :\mathbf {s} }}-{\sqrt {\cfrac {2}{3}}}~\sigma _{y}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1cab8e2de0fbbde1c4f08374a4e2a23be001654)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/843d3c19bf46a710c9c903d6fab46c2b4b6be65e)
The deviatoric part of
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50bb88b9b07d1c7c9e1476fa91fd5bd526955e63)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07377949583c4c1e2ddfe402541b8818d5a30ef1)
Now,
![{\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\mathit {1}}}=\sigma _{ij}\delta _{ij}=\sigma _{ii}={\text{tr}}({\boldsymbol {\sigma }})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76e48ef2068556e79e310d84a02be96704588845)
and
![{\displaystyle {\boldsymbol {\mathit {1}}}:{\boldsymbol {\mathit {1}}}=\delta _{ij}\delta _{ij}=\delta _{ii}=3~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f82151902b080c68399f7d85d9eea8890a6cec3)
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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b88e39a90318e41dfd00132b5a998a03e75fe6b8)
The derivative with respect to
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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d69c30721bdd2926c43ef55d4af01ad9646dafb0)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/609efa489a5313108fb9d744dfe0dff56375a19a)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ccb9645cfe70047db9d157bea58ef28147db8ad6)
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 }}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e55841f2c3cb8b86e7cdc2761902bdf00659adc6)
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})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/caf0b9376de81dfb3be2de2a1867b903a92d7b8a)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecce5c59cc50cca7df15f1f0f4b0f1c4f455108f)
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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2676cf7733e8e6b73e06af8796fe515d5115a8d9)
Plugging the above results into the expression for the derivative
of
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} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7e9217cedfc554469400c0236c0a6e7bfb39783)
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} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d3847a833309485f51c7d41f54ee2632ee36df0)
The required expression is
![{\displaystyle {\frac {\partial f}{\partial {\boldsymbol {\sigma }}}}=f_{\boldsymbol {\sigma }}={\sqrt {\cfrac {3}{2}}}~\mathbf {n} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7aafa5aa7fa844e84c35d03580fc80a9b2294234)