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

$\dot{\mathit{\sigma }}={𝖢}^{\text{ep}}:\dot{\mathit{\epsilon }}.$

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

$\dot{\mathit{\sigma }}=𝖢:(\dot{\mathit{\epsilon }}-\dot{\gamma }{f}_{\mathit{\sigma }}).$

From the previous part

$\dot{\gamma }=\frac{f_{\mathit{\sigma }}:𝖢:\dot{\mathit{\epsilon }}}{f_{\mathit{\sigma }}:𝖢:f_{\mathit{\sigma }}-\sqrt{\frac{2}{3}}{f}_{\alpha }\frac{{\mathit{\epsilon }}^p:f_{\mathit{\sigma }}}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}-\frac{\chi }{\rho C_p}{f}_T\mathit{\sigma }:f_{\mathit{\sigma }}}.$

Plug in expression for stress rate to get

$\begin{array}{cc}\dot{\mathit{\sigma }}& =𝖢:(\dot{\mathit{\epsilon }}-\frac{f_{\mathit{\sigma }}:𝖢:\dot{\mathit{\epsilon }}}{f_{\mathit{\sigma }}:𝖢:f_{\mathit{\sigma }}-\sqrt{\frac{2}{3}}{f}_{\alpha }\frac{{\mathit{\epsilon }}^p:f_{\mathit{\sigma }}}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}-\frac{\chi }{\rho C_p}{f}_T\mathit{\sigma }:f_{\mathit{\sigma }}}{f}_{\mathit{\sigma }})\\ & =𝖢:\dot{\mathit{\epsilon }}-\frac{f_{\mathit{\sigma }}:𝖢:\dot{\mathit{\epsilon }}}{f_{\mathit{\sigma }}:𝖢:f_{\mathit{\sigma }}-\sqrt{\frac{2}{3}}{f}_{\alpha }\frac{{\mathit{\epsilon }}^p:f_{\mathit{\sigma }}}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}-\frac{\chi }{\rho C_p}{f}_T\mathit{\sigma }:f_{\mathit{\sigma }}}𝖢:f_{\mathit{\sigma }}\end{array}$

We have to express the above in the form

$\dot{\mathit{\sigma }}={𝖢}^{\text{ep}}:\dot{\mathit{\epsilon }}⟹{\dot{\sigma }}_{ij}=C_{ijkl}^{\text{ep}}{\dot{\epsilon }}_{kl}.$

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

$\dot{\mathit{\sigma }}=𝖢:\dot{\mathit{\epsilon }}-\left(\frac{f_{\mathit{\sigma }}:𝖢:\dot{\mathit{\epsilon }}}{\text{denom.}}\right)𝖢:f_{\mathit{\sigma }}$

In index notation, we can write

${\dot{\sigma }}_{ij}=C_{ijkl}{\dot{\epsilon }}_{kl}-\frac{f_{pq}^{\mathit{\sigma }}{C}_{pqrs}{\dot{\epsilon }}_{rs}}{\text{denom.}}{C}_{ijkl}{f}_{kl}^{\mathit{\sigma }}$

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

$\begin{array}{cc}{f}_{pq}^{\mathit{\sigma }}{C}_{pqrs}{\dot{\epsilon }}_{rs}{C}_{ijkl}{f}_{kl}^{\mathit{\sigma }}& =\left(C_{pqrs}{f}_{pq}^{\mathit{\sigma }}\right)\left(C_{ijkl}{f}_{kl}^{\mathit{\sigma }}\right){\dot{\epsilon }}_{rs}\\ & =\left(C_{rspq}{f}_{pq}^{\mathit{\sigma }}\right)\left(C_{ijkl}{f}_{kl}^{\mathit{\sigma }}\right){\dot{\epsilon }}_{rs}\text{Major symmetry of}𝖢⟹C_{pqrs}=C_{rspq}\\ & =\left(C_{ijkl}{f}_{kl}^{\mathit{\sigma }}\right)\left(C_{rspq}{f}_{pq}^{\mathit{\sigma }}\right){\dot{\epsilon }}_{rs}\\ & \equiv A_{ij}{B}_{rs}{\dot{\epsilon }}_{rs}\equiv M_{ijrs}{\dot{\epsilon }}_{rs}\\ & =𝖬:\dot{\mathit{\epsilon }}.\end{array}$

In the above

$M_{ijrs}=A_{ij}{B}_{rs}⟹𝖬=𝑨\otimes 𝑩$

and

$\begin{array}{cc}{A}_{ij}& =C_{ijkl}{f}_{kl}^{\mathit{\sigma }}⟹𝑨=𝖢:f_{\mathit{\sigma }}\\ B_{rs}& =C_{rspq}{f}_{pq}^{\mathit{\sigma }}⟹𝑩=𝖢:f_{\mathit{\sigma }}\end{array}$

Therefore,

$𝖬=(𝖢:f_{\mathit{\sigma }})\otimes (𝖢:f_{\mathit{\sigma })}.$

This gives us

$\dot{\mathit{\sigma }}=𝖢:\dot{\mathit{\epsilon }}-\left(\frac{(𝖢:f_{\mathit{\sigma }})\otimes (𝖢:f_{\mathit{\sigma })}}{\text{denom.}}\right):\dot{\mathit{\epsilon }}$

or,

$\dot{\mathit{\sigma }}=[𝖢-\left(\frac{(𝖢:f_{\mathit{\sigma }})\otimes (𝖢:f_{\mathit{\sigma })}}{\text{denom.}}\right)]:\dot{\mathit{\epsilon }}.$

Hence

${𝖢}^{\text{ep}}=𝖢-\left(\frac{(𝖢:f_{\mathit{\sigma }})\otimes (𝖢:f_{\mathit{\sigma })}}{\text{denom.}}\right).$

The continuum elastic-plastic tangent modulus is therefore

${𝖢}^{\text{ep}}=𝖢-\left(\frac{(𝖢:f_{\mathit{\sigma }})\otimes (𝖢:f_{\mathit{\sigma })}}{f_{\mathit{\sigma }}:𝖢:f_{\mathit{\sigma }}-\sqrt{\frac{2}{3}}{f}_{\alpha }\frac{{\mathit{\epsilon }}^p:f_{\mathit{\sigma }}}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}-\frac{\chi }{\rho C_p}{f}_T\mathit{\sigma }:f_{\mathit{\sigma }}}\right).$