Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 9

Problem 1: Part 9: Elastic-plastic tangent modulus
Assume that the elastic response of the material is linear, i.e.,

\boldsymbol{\mathsf{C}} = \lambda\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2 \mu\boldsymbol{\mathsf{I}} ~. $$

Derive the expression for the elastic-plastic tangent modulus for a von Mises yield condition with Johnson-Cook flow stress for a linear elastic material using the expressions that you have derived in the previous parts.

The elastic-plastic tangent modulus is given by

\boldsymbol{\mathsf{C}}^{\text{ep}} = \boldsymbol{\mathsf{C}} - \left(\cfrac{(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf

{C}}:f_{\boldsymbol{\sigma})}} {f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}}\right)~. $$ From the previous parts

f_{\boldsymbol{\sigma}} = \sqrt{\cfrac{3}{2}}~\mathbf{n} ~; f_{\alpha} = - n~B~\alpha^{n-1} \left[1 - \left(\cfrac{T - T_0}{T_m -T_0}\right)\right] ~; f_T = \left(\cfrac{1}{T_m - T_0}\right) \left[\sigma_0 + B \alpha^n\right] ~. $$ Therefore,

\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} = \sqrt{\cfrac{3}{2}}~(\lambda~\boldsymbol

{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf{I}}):\mathbf{n} = \sqrt{\cfrac{3}{2}}~(\lambda~\text{tr}(\mathbf{n})~\boldsymbol{\mathit{1}} +

2~\mu~\mathbf{n}) = \sqrt{\cfrac{3}{2}}~2~\mu~\mathbf{n} ~. $$

Some of the results used in the above derivation are shown below.

Recall (from previous homework):

\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}}:\mathbf{n} = \delta_{ij}~\delta_{kl}

~n_{kl} = n_{kk}~\delta_{ij} = \text{tr}(\mathbf{n})~\boldsymbol{\mathit{1}} $$ and

\boldsymbol{\mathsf{I}}:\mathbf{n} = \frac{1}{2}~(\delta_{ik}~\delta_{jl} + \delta_{il}

~\delta_{jk}) n_{kl} = \frac{1}{2}~(\delta_{jl}~n_{il} + \delta_{jk}~n_{ki}) = \frac{1}{2}~(n_{ij} + n_{ji}) = n_{ij} = \mathbf{n} $$ (we have used the symmetry of the stress tensor above.)

Also,

\text{tr}(\mathbf{n}) = \text{tr}\left(\cfrac{\mathbf{s}}{\lVert\mathbf{s}\rVert_{}}\right) = \cfrac{\text{tr}(\mathbf{s})}{\lVert\mathbf{s}\rVert_{}} $$ Now,

\mathbf{s} = s_{ij} = \sigma_{ij} - \frac{1}{3}~\sigma_{kk}~\delta_{ij} \qquad \implies \qquad \text{tr}(\mathbf{s}) = s_{ii} = \sigma_{ii} - \frac{1}{3}~\sigma_{kk}~\delta_{ii} = \sigma_{ii} - \frac{1}{3}~\sigma_{kk}~3 = \sigma_{ii} - \sigma_{kk} = 0 $$ Therefore,

\text{tr}(\mathbf{n}) = 0 ~. $$

Hence,

(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf{C}}:f_

{\boldsymbol{\sigma})} = \cfrac{3}{2}~4~\mu^2~\mathbf{n}\otimes\mathbf{n} = 6~\mu^2~\mathbf{n}\otimes\mathbf{n} $$ and

f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} = \cfrac{3}{2}

~2~\mu~\mathbf{n}:\mathbf{n} = 3~\mu~\left(\cfrac{\mathbf{s}}{\sqrt{\mathbf{s}:\mathbf{s}}}\right): \left(\cfrac{\mathbf{s}}{\sqrt{\mathbf{s}:\mathbf{s}}}\right) = 3~\mu~\cfrac{\mathbf{s}:\mathbf{s}}{\mathbf{s}:\mathbf{s}} = 3~\mu ~. $$ Plugging in expression for $$\boldsymbol{\mathsf{C}}^{\text{ep}}$$ we get

\boldsymbol{\mathsf{C}}^{\text{ep}} = \lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}} - \left(\cfrac{ 6~\mu^2~\mathbf{n}\otimes\mathbf{n} } {3~\mu -n~B~\alpha^{n-1}\left[1 - \left(\cfrac{T-T_0}{T_m-T_0}\right)\right] ~\cfrac{\boldsymbol{\varepsilon}^p:\mathbf{n}}{\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \sqrt{\cfrac{3}{2}}~\cfrac{\chi}{\rho~C_p}~ \left(\cfrac{1}{T_m - T_0}\right)\left[\sigma_0 + B \alpha^n\right] ~\boldsymbol{\sigma}:\mathbf{n}}\right)~. $$ Now,

\boldsymbol{\sigma}:\mathbf{n} = \cfrac{(\mathbf{s} + \frac{1}{3}~\text{tr}(\boldsymbol

{\sigma})~\boldsymbol{\mathit{1}}):\mathbf{s}}{\sqrt{\mathbf{s}:\mathbf{s}}} = \cfrac{\mathbf{s}:\mathbf{s}}{\sqrt{\mathbf{s}:\mathbf{s}}} + \frac{1}{3}~\text{tr}(\boldsymbol{\sigma})~\cfrac{\boldsymbol{\mathit{1}}:\mathbf{s}}{\sqrt

{\mathbf{s}:\mathbf{s}}} = \sqrt{\mathbf{s}:\mathbf{s}} + \frac{1}{3}~\text{tr}(\boldsymbol{\sigma})~\cfrac{\text{tr}(\mathbf{s})}{\sqrt{\mathbf

{s}:\mathbf{s}}} = \sqrt{\mathbf{s}:\mathbf{s}} = \lVert\mathbf{s}\rVert_{} $$ Therefore, the elastic-plastic tangent modulus can be written as

\boldsymbol{\mathsf{C}}^{\text{ep}} = \lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\left[\boldsymbol

{\mathsf{I}} - \cfrac{ 3~\mu~\mathbf{n}\otimes\mathbf{n} } {3~\mu -n~B~\alpha^{n-1}\left[1 - \left(\cfrac{T-T_0}{T_m-T_0}\right)\right] ~\cfrac{\boldsymbol{\varepsilon}^p:\mathbf{n}}{\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \sqrt{\cfrac{3}{2}}~\cfrac{\chi}{\rho~C_p}~ \left(\cfrac{1}{T_m - T_0}\right)\left[\sigma_0 + B \alpha^n\right] ~\lVert\mathbf{s}\rVert_{}}\right]~. $$