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.,

$𝖢=\lambda 1\otimes 1+2\mu 𝖨.$

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

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

From the previous parts

$f_{\mathit{\sigma }}=\sqrt{\frac{3}{2}}𝐧;f_{\alpha }=-nB{\alpha }^{n-1}[1-\left(\frac{T-T_0}{T_m-T_0}\right)];f_T=\left(\frac{1}{T_m-T_0}\right)[{\sigma }_0+B{\alpha }^n].$

Therefore,

$𝖢:f_{\mathit{\sigma }}=\sqrt{\frac{3}{2}}(\lambda 1\otimes 1+2\mu 𝖨):𝐧=\sqrt{\frac{3}{2}}(\lambda \text{tr}(𝐧)1+2\mu 𝐧)=\sqrt{\frac{3}{2}}2\mu 𝐧.$

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

Recall (from previous homework):

$1\otimes 1:𝐧={\delta }_{ij}{\delta }_{kl}{n}_{kl}=n_{kk}{\delta }_{ij}=\text{tr}(𝐧)1$

and

$𝖨:𝐧=\frac{1}{2}({\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk})n_{kl}=\frac{1}{2}({\delta }_{jl}{n}_{il}+{\delta }_{jk}{n}_{ki})=\frac{1}{2}(n_{ij}+n_{ji})=n_{ij}=𝐧$

(we have used the symmetry of the stress tensor above.)

Also,

$\text{tr}(𝐧)=\text{tr}\left(\frac{𝐬}{\Vert 𝐬{\Vert }_{}}\right)=\frac{\text{tr}(𝐬)}{\Vert 𝐬{\Vert }_{}}$

Now,

$𝐬=s_{ij}={\sigma }_{ij}-\frac{1}{3}{\sigma }_{kk}{\delta }_{ij}⟹\text{tr}(𝐬)=s_{ii}={\sigma }_{ii}-\frac{1}{3}{\sigma }_{kk}{\delta }_{ii}={\sigma }_{ii}-\frac{1}{3}{\sigma }_{kk}3={\sigma }_{ii}-{\sigma }_{kk}=0$

Therefore,

$\text{tr}(𝐧)=0.$

Hence,

$(𝖢:f_{\mathit{\sigma }})\otimes (𝖢:f_{\mathit{\sigma })}=\frac{3}{2}4{\mu }^2𝐧\otimes 𝐧=6{\mu }^2𝐧\otimes 𝐧$

and

$f_{\mathit{\sigma }}:𝖢:f_{\mathit{\sigma }}=\frac{3}{2}2\mu 𝐧:𝐧=3\mu \left(\frac{𝐬}{\sqrt{𝐬:𝐬}}\right):\left(\frac{𝐬}{\sqrt{𝐬:𝐬}}\right)=3\mu \frac{𝐬:𝐬}{𝐬:𝐬}=3\mu .$

Plugging in expression for ${𝖢}^{\text{ep}}$ we get

${𝖢}^{\text{ep}}=\lambda 1\otimes 1+2\mu 𝖨-\left(\frac{6{\mu }^2𝐧\otimes 𝐧}{3\mu -nB{\alpha }^{n-1}[1-\left(\frac{T-T_0}{T_m-T_0}\right)]\frac{{\mathit{\epsilon }}^p:𝐧}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}-\sqrt{\frac{3}{2}}\frac{\chi }{\rho C_p}\left(\frac{1}{T_m-T_0}\right)[{\sigma }_0+B{\alpha }^n]\mathit{\sigma }:𝐧}\right).$

Now,

$\mathit{\sigma }:𝐧=\frac{(𝐬+\frac{1}{3}\text{tr}(\mathit{\sigma })1):𝐬}{\sqrt{𝐬:𝐬}}=\frac{𝐬:𝐬}{\sqrt{𝐬:𝐬}}+\frac{1}{3}\text{tr}(\mathit{\sigma })\frac{1:𝐬}{\sqrt{𝐬:𝐬}}=\sqrt{𝐬:𝐬}+\frac{1}{3}\text{tr}(\mathit{\sigma })\frac{\text{tr}(𝐬)}{\sqrt{𝐬:𝐬}}=\sqrt{𝐬:𝐬}=\Vert 𝐬{\Vert }_{}$

Therefore, the elastic-plastic tangent modulus can be written as

${𝖢}^{\text{ep}}=\lambda 1\otimes 1+2\mu [𝖨-\frac{3\mu 𝐧\otimes 𝐧}{3\mu -nB{\alpha }^{n-1}[1-\left(\frac{T-T_0}{T_m-T_0}\right)]\frac{{\mathit{\epsilon }}^p:𝐧}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}-\sqrt{\frac{3}{2}}\frac{\chi }{\rho C_p}\left(\frac{1}{T_m-T_0}\right)[{\sigma }_0+B{\alpha }^n]\Vert 𝐬{\Vert }_{}}].$