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

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