Show that
![{\displaystyle \mathbf {s} _{n+1}=\mathbf {s} _{n+1}^{\text{trial}}-2~\mu ~\Delta \gamma ~\mathbf {n} _{n}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58f781ed975cb658000179342ddeac5bd1648aa1)
We have
![{\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}_{n+1}&={\boldsymbol {\mathsf {C}}}:{\boldsymbol {\varepsilon }}_{n+1}^{e}={\boldsymbol {\mathsf {C}}}:({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n+1}^{p})\\&=(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n+1}^{p})\\&=(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n})\\&=\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})~{\boldsymbol {\mathit {1}}}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~{\text{tr}}(\mathbf {n} _{n})~{\boldsymbol {\mathit {1}}}\right]+2~\mu ~\left[{\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\right]\\&=\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~{\text{tr}}(\mathbf {n} _{n})\right]{\boldsymbol {\mathit {1}}}+2~\mu ~\left[{\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\right]\\&=\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\right]{\boldsymbol {\mathit {1}}}+2~\mu ~\left[{\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\right]\qquad ~({\text{since}}~{\text{tr}}(\mathbf {n} )=0)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbff3be883bc91e7943c72e63a2bfcf1e884368e)
Now
![{\displaystyle \mathbf {s} _{n+1}={\boldsymbol {\sigma }}_{n+1}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }}_{n+1})~{\boldsymbol {\mathit {1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5452e48e4c7a5f804d49be8cfea88e9ecf7717f)
The trace of the stress is given by
![{\displaystyle {\begin{aligned}{\text{tr}}({\boldsymbol {\sigma }}_{n+1})&=\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\right]{\text{tr}}({\boldsymbol {\mathit {1}}})+2~\mu ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~{\text{tr}}(\mathbf {n} _{n})\right]\\&=3~\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\right]+2~\mu ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\right]\qquad ~({\text{since}}~{\text{tr}}(\mathbf {n} )=0)\\&=(3~\lambda +2~\mu ){\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-(3~\lambda +2~\mu ){\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c064a5b4677e61bd63bb928295125fb551aa40d)
Therefore,
![{\displaystyle {\begin{aligned}\mathbf {s} _{n+1}&={\boldsymbol {\sigma }}_{n+1}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }}_{n+1})~{\boldsymbol {\mathit {1}}}\\&=\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\right]{\boldsymbol {\mathit {1}}}+2~\mu ~\left[{\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\right]\\&\qquad -(\lambda +{\cfrac {2}{3}}~\mu ){\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}+(\lambda +{\cfrac {2}{3}}~\mu ){\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})~{\boldsymbol {\mathit {1}}}\\&=2~\mu \left[{\boldsymbol {\varepsilon }}_{n+1}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}\right]-2~\mu \left[{\boldsymbol {\varepsilon }}_{n}^{p}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})~{\boldsymbol {\mathit {1}}}\right]-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\\&=2~\mu \mathbf {e} _{n+1}-2~\mu \left[{\boldsymbol {\varepsilon }}_{n}-{\boldsymbol {\varepsilon }}_{n}^{e}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}-{\boldsymbol {\varepsilon }}_{n}^{e})~{\boldsymbol {\mathit {1}}}\right]-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\\&=2~\mu \mathbf {e} _{n+1}-2~\mu \left[{\boldsymbol {\varepsilon }}_{n}-{\boldsymbol {\varepsilon }}_{n}^{e}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n})~{\boldsymbol {\mathit {1}}}+{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{e})~{\boldsymbol {\mathit {1}}}\right]-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\\&=2~\mu \left[\mathbf {e} _{n+1}-\mathbf {e} _{n}\right]+2~\mu \left[{\boldsymbol {\varepsilon }}_{n}^{e}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{e})~{\boldsymbol {\mathit {1}}}\right]-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\\&=\mathbf {s} _{n+1}^{\text{trial}}-\mathbf {s} _{n}+2~\mu \left[{\boldsymbol {\varepsilon }}_{n}^{e}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{e})~{\boldsymbol {\mathit {1}}}\right]-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/941f7ba6d41d02ccb5622ef96857e94b4973f309)
The stress-strain relation is
![{\displaystyle {\boldsymbol {\sigma }}_{n}=\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{e})~{\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\varepsilon }}_{n}^{e}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ad713b8ce22e3da6f4b52ad64cc899ff4395609)
Hence,
![{\displaystyle {\begin{aligned}\mathbf {s} _{n}&=\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{e})~{\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\varepsilon }}_{n}^{e}-{\frac {1}{3}}~\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{e})~{\text{tr}}({\boldsymbol {\mathit {1}}})~{\boldsymbol {\mathit {1}}}-{\cfrac {2}{3}}~\mu ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{e})~{\boldsymbol {\mathit {1}}}\\&=2~\mu \left[{\boldsymbol {\varepsilon }}_{n}^{e}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{e})~{\boldsymbol {\mathit {1}}}\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4d7426dd91bbcd3f683368adab682691bbdc6ad)
Plugging into expression for
, we get
![{\displaystyle \mathbf {s} _{n+1}=\mathbf {s} _{n+1}^{\text{trial}}-\mathbf {s} _{n}+\mathbf {s} _{n}-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4c8c6ca4a22f0607876dc955a72418508167bcd)
Therefore,
![{\displaystyle {\mathbf {s} _{n+1}=\mathbf {s} _{n+1}^{\text{trial}}-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22bb7f0bd73d27be8ac9b1fefb89137ec7ad4cf9)
Remark: If we 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)
then the above equation takes the form
![{\displaystyle {\mathbf {s} _{n+1}=\mathbf {s} _{n+1}^{\text{trial}}-2~\mu ~\Delta \gamma ~\mathbf {n} _{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14c5fedbc2d105476aef7a644ac55ca1f1d863c1)
These are equivalent.