Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 15

Problem 1: Part 15: Finding the plastic flow parameter
The discretized form of the Kuhn-Tucker conditions in conjunction with the consistency condition gives us

f(\boldsymbol{\sigma}_{n+1}, \alpha_{n+1}, T_{n+1}) = 0 ~. $$ Use this condition and the relations you have derived in the previous sections to arrive at a nonlinear equation in $$\Delta\gamma$$ that can be solved using Newton iterations.

The yield function is

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

\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,

\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,

\mathbf{s}_{n+1} = \mathbf{s}_{n+1}^{\text{trial}} - 2~\mu~\sqrt{\cfrac{3}{2}}

~\Delta\gamma~\mathbf{n}_n $$ Therefore,

\begin{align} \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{align} $$ Plugging into the discretized yield condition, we have

\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,

\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

\begin{align} \alpha_{n+1}& = \alpha_n + \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol {\varepsilon}^p_n\rVert_{}}\\ T_{n+1} & = T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} ~\lVert\mathbf{s}_n\rVert_{} \end{align} $$ Therefore,

\left[\sigma_0 + B \left\{\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\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 $$ The nonlinear equation in $$\Delta\gamma$$ is

{ \begin{align} 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}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\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{align} } $$