Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 6

Problem 1: Part 6: Continuum elastic-plastic tangent modulus

The continuum elastic-plastic tangent modulus is defined by the following relation

$\dot{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}}^{\text{ep}}:\dot{\boldsymbol {\varepsilon}} ~.$

Derive an expression for the elastic plastic tangent modulus using the results you have derived in the previous parts.

The stress rate is given by

$\dot{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}} : \left(\dot{\boldsymbol{\varepsilon}} - \dot{\gamma}f_{\boldsymbol {\sigma}}\right)~.$

From the previous part

$\dot{\gamma} = \cfrac{f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\dot{\boldsymbol{\varepsilon}}}{ f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}} ~.$

Plug in expression for stress rate to get

$\begin{align} \dot{\boldsymbol{\sigma}} & = \boldsymbol{\mathsf{C}} : \left(\dot{\boldsymbol{\varepsilon}} - \cfrac{f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\dot{\boldsymbol{\varepsilon}}}{ f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}}~ f_{\boldsymbol{\sigma}}\right) \\ & =\boldsymbol{\mathsf{C}} : \dot{\boldsymbol{\varepsilon}} - \cfrac{f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\dot{\boldsymbol{\varepsilon}}}{ f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}}~ \boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} \end{align}$

We have to express the above in the form

$\dot{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}}^{\text{ep}}:\dot{\boldsymbol {\varepsilon}} \qquad\implies\qquad \dot{\sigma}_{ij} = C^{\text{ep}}_{ijkl}~\dot{\varepsilon}_{kl} ~.$

Since the denominator is a scalar, we don't have to worry about it for this calculation. In that case we can write

$\dot{\boldsymbol{\sigma}}=\boldsymbol{\mathsf{C}} : \dot{\boldsymbol{\varepsilon}} - \left(\cfrac{f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\dot{\boldsymbol {\varepsilon}}}{\text{denom.}}\right) \boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}}$

In index notation, we can write

$\dot{\sigma}_{ij} = C_{ijkl}~\dot{\varepsilon}_{kl} - \cfrac{f^{\boldsymbol{\sigma}}_{pq}~C_{pqrs}~\dot{\varepsilon}_{rs}}{\text{denom.}}~ C_{ijkl}~f^{\boldsymbol{\sigma}}_{kl}$

Let us manipulate the numerator of the second term above so that we get what we need. Thus

$\begin{align} f^{\boldsymbol{\sigma}}_{pq}~C_{pqrs}~\dot{\varepsilon}_{rs}~C_{ijkl}~f^{\boldsymbol {\sigma}}_{kl} & = \left(C_{pqrs}~f^{\boldsymbol{\sigma}}_{pq}\right)~ \left(C_{ijkl}~f^{\boldsymbol{\sigma}}_{kl}\right)~\dot{\varepsilon}_{rs} \\ & = \left(C_{rspq}~f^{\boldsymbol{\sigma}}_{pq}\right)~ \left(C_{ijkl}~f^{\boldsymbol{\sigma}}_{kl}\right)~\dot{\varepsilon}_{rs} \qquad \text{Major symmetry of}~ \boldsymbol{\mathsf{C}} \implies C_{pqrs} = C_{rspq} \\ & = \left(C_{ijkl}~f^{\boldsymbol{\sigma}}_{kl}\right)~ \left(C_{rspq}~f^{\boldsymbol{\sigma}}_{pq}\right)~ ~\dot{\varepsilon}_{rs} \\ & \equiv A_{ij}~B_{rs}~\dot{\varepsilon}_{rs} \equiv M_{ijrs}~\dot{\varepsilon}_{rs} \\ & = \boldsymbol{\mathsf{M}}:\dot{\boldsymbol{\varepsilon}} ~. \end{align}$

In the above

$M_{ijrs} = A_{ij}~B_{rs} \qquad\implies\qquad \boldsymbol{\mathsf{M}} = \boldsymbol{A} \otimes\boldsymbol{B}$

and

$\begin{align} A_{ij} & = C_{ijkl}~f^{\boldsymbol{\sigma}}_{kl} \qquad\implies\qquad \boldsymbol{A} = \boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} \\ B_{rs} & = C_{rspq}~f^{\boldsymbol{\sigma}}_{pq} \qquad\implies\qquad \boldsymbol{B} = \boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} \end{align}$

Therefore,

$\boldsymbol{\mathsf{M}} = (\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes (\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma})} ~.$

This gives us

$\dot{\boldsymbol{\sigma}}=\boldsymbol{\mathsf{C}} : \dot{\boldsymbol{\varepsilon}} - \left(\cfrac{(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf {C}}:f_{\boldsymbol{\sigma})}} {\text{denom.}}\right): \dot{\boldsymbol{\varepsilon}}$

or,

$\dot{\boldsymbol{\sigma}} = \left[\boldsymbol{\mathsf{C}} - \left(\cfrac{(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf {C}}:f_{\boldsymbol{\sigma})}} {\text{denom.}}\right)\right]: \dot{\boldsymbol{\varepsilon}} ~.$

Hence

$\boldsymbol{\mathsf{C}}^{\text{ep}} = \boldsymbol{\mathsf{C}} - \left(\cfrac{(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf {C}}:f_{\boldsymbol{\sigma})}} {\text{denom.}}\right) ~.$

The continuum elastic-plastic tangent modulus is therefore

${ \boldsymbol{\mathsf{C}}^{\text{ep}} = \boldsymbol{\mathsf{C}} - \left(\cfrac{(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf {C}}:f_{\boldsymbol{\sigma})}} {f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}}\right)~. }$