Problem 1: Part 15: Finding the plastic flow parameter
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The discretized form of the Kuhn-Tucker conditions in conjunction
with the consistency condition gives us
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{\displaystyle f({\boldsymbol {\sigma }}_{n+1},\alpha _{n+1},T_{n+1})=0~.}
Use this condition and the relations you have derived in the previous sections to arrive at
a nonlinear equation in
Δ
γ
{\displaystyle \Delta \gamma }
The yield function is
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{\displaystyle f={\sqrt {\cfrac {3}{2}}}~{\sqrt {\mathbf {s} :\mathbf {s} }}-\left[\sigma _{0}+B\alpha ^{n}\right]\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]}
Therefore the discretized form of Kuhn-Tucker + consistency is
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{\displaystyle {\sqrt {\cfrac {3}{2}}}~{\sqrt {\mathbf {s} _{n+1}:\mathbf {s} _{n+1}}}-\left[\sigma _{0}+B\alpha _{n+1}^{n}\right]\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]=0}
or,
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{\displaystyle \mathbf {s} _{n+1}:\mathbf {s} _{n+1}={\cfrac {2}{3}}\left[\sigma _{0}+B\alpha _{n+1}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}}
Now,
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{\displaystyle \mathbf {s} _{n+1}=\mathbf {s} _{n+1}^{\text{trial}}-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}}
Therefore,
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{\displaystyle {\begin{aligned}\mathbf {s} _{n+1}:\mathbf {s} _{n+1}&=\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-4~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+4~\mu ^{2}~{\cfrac {3}{2}}~(\Delta \gamma )^{2}~\mathbf {n} _{n}:\mathbf {n} _{n}\\&=\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-4~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+6~\mu ^{2}~(\Delta \gamma )^{2}\end{aligned}}}
Plugging into the discretized yield condition, we have
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{\displaystyle {\cfrac {2}{3}}\left[\sigma _{0}+B\alpha _{n+1}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}=\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-4~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+6~\mu ^{2}~(\Delta \gamma )^{2}}
or,
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{\displaystyle \left[\sigma _{0}+B\alpha _{n+1}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}={\cfrac {3}{2}}~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-6~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+9~\mu ^{2}~(\Delta \gamma )^{2}}
Also
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{\displaystyle {\begin{aligned}\alpha _{n+1}&=\alpha _{n}+\Delta \gamma ~{\cfrac {{\boldsymbol {\varepsilon }}_{n}^{p}:\mathbf {n} _{n}}{\lVert {\boldsymbol {\varepsilon }}_{n}^{p}\rVert _{}}}\\T_{n+1}&=T_{n}+{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi ~\Delta \gamma }{\rho _{n}~C_{p}}}~\lVert \mathbf {s} _{n}\rVert _{}\end{aligned}}}
Therefore,
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{\displaystyle \left[\sigma _{0}+B\left\{\alpha _{n}+\Delta \gamma ~{\cfrac {{\boldsymbol {\varepsilon }}_{n}^{p}:\mathbf {n} _{n}}{\lVert {\boldsymbol {\varepsilon }}_{n}^{p}\rVert _{}}}\right\}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n}+{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi ~\Delta \gamma }{\rho _{n}~C_{p}}}~\lVert \mathbf {s} _{n}\rVert _{}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}={\cfrac {3}{2}}~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-6~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+9~\mu ^{2}~(\Delta \gamma )^{2}}
The nonlinear equation in
Δ
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{\displaystyle \Delta \gamma }
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{\displaystyle {\begin{aligned}g(\Delta \gamma )&=0\\&=9~\mu ^{2}~(\Delta \gamma )^{2}-6~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}\\&\qquad -\left[\sigma _{0}+B\left\{\alpha _{n}+\Delta \gamma ~{\cfrac {{\boldsymbol {\varepsilon }}_{n}^{p}:\mathbf {n} _{n}}{\lVert {\boldsymbol {\varepsilon }}_{n}^{p}\rVert _{}}}\right\}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n}+{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi ~\Delta \gamma }{\rho _{n}~C_{p}}}\lVert \mathbf {s} _{n}\rVert _{}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}\\&\qquad +{\cfrac {3}{2}}~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}\end{aligned}}}
![{\displaystyle f={\sqrt {\cfrac {3}{2}}}~{\sqrt {\mathbf {s} :\mathbf {s} }}-\left[\sigma _{0}+B\alpha ^{n}\right]\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/652d92be8a254d66197b08533f5723447f38c1c5)
![{\displaystyle {\sqrt {\cfrac {3}{2}}}~{\sqrt {\mathbf {s} _{n+1}:\mathbf {s} _{n+1}}}-\left[\sigma _{0}+B\alpha _{n+1}^{n}\right]\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac618039e6209cf5b043b4b5ca5e4439c801bbe9)
![{\displaystyle \mathbf {s} _{n+1}:\mathbf {s} _{n+1}={\cfrac {2}{3}}\left[\sigma _{0}+B\alpha _{n+1}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c07af15a0080e2f12acaa279b9f67eec441df289)
![{\displaystyle {\cfrac {2}{3}}\left[\sigma _{0}+B\alpha _{n+1}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}=\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-4~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+6~\mu ^{2}~(\Delta \gamma )^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc057498ff03921a79e5d72ac2645d856c328a47)
![{\displaystyle \left[\sigma _{0}+B\alpha _{n+1}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}={\cfrac {3}{2}}~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-6~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+9~\mu ^{2}~(\Delta \gamma )^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63c6c92a153b8d159468f91cbc3de0f5c6d17427)
![{\displaystyle \left[\sigma _{0}+B\left\{\alpha _{n}+\Delta \gamma ~{\cfrac {{\boldsymbol {\varepsilon }}_{n}^{p}:\mathbf {n} _{n}}{\lVert {\boldsymbol {\varepsilon }}_{n}^{p}\rVert _{}}}\right\}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n}+{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi ~\Delta \gamma }{\rho _{n}~C_{p}}}~\lVert \mathbf {s} _{n}\rVert _{}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}={\cfrac {3}{2}}~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-6~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+9~\mu ^{2}~(\Delta \gamma )^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e35e2061271dc3e5440613ef624307f2c4b83010)
![{\displaystyle {\begin{aligned}g(\Delta \gamma )&=0\\&=9~\mu ^{2}~(\Delta \gamma )^{2}-6~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}\\&\qquad -\left[\sigma _{0}+B\left\{\alpha _{n}+\Delta \gamma ~{\cfrac {{\boldsymbol {\varepsilon }}_{n}^{p}:\mathbf {n} _{n}}{\lVert {\boldsymbol {\varepsilon }}_{n}^{p}\rVert _{}}}\right\}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n}+{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi ~\Delta \gamma }{\rho _{n}~C_{p}}}\lVert \mathbf {s} _{n}\rVert _{}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}\\&\qquad +{\cfrac {3}{2}}~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5112828216e69e68f98cf5ebbfc2cc792246ca5)