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I am trying to find material stiffness matrix for linear elasticity for finite element code,

$$\mathbf{\sigma} = \lambda \hspace{1pt} \operatorname{tr}{\left(\mathbf{\epsilon}\right)}+ 2\mu\mathbf{\epsilon} \,,$$

where $\mathbf{\sigma,\epsilon}$ are Cauchy stress and corresponding conjugate strain respectively; both being second order tensors. $\operatorname{tr}\left(\mathbf{\epsilon} \right)$ is trace of the tensor.

I want to find

$$\frac{\partial \mathbf{\sigma}}{\partial\mathbf{\epsilon}}$$

Considering stress and strain as $6 {\times} 1$ column vectors, I started like this:

$$ \begin{align} \mathbb{C_{ij}}& =\frac{\partial {\sigma_{i}}}{\partial{\epsilon_{j}}} \\[2.5px] & = \frac{\partial}{\partial \epsilon_{j}} \left(\lambda\epsilon_{i} + 2\mu\epsilon_{i} \right) \\[2.5px] & =\lambda\frac{\partial}{\partial \epsilon_{j}}\epsilon_{i} + 2\mu\frac{\partial}{\partial \epsilon_{j}}\epsilon_{i} \\[2.5px] & =\left( \lambda + 2 \mu \right)\delta_{ij} \end{align} $$

However, $\mathbb{C_{ij}}$ is given differently – as 6x6 matrix in here.

(Note: The above equations are in incremental form in reality, but I just avoided that notation.)

Can someone comment what is wrong in my derivation and what material stiffness matrix is right to use?

Edit:

After answer from Chemomechanics, I understood the problem, and rewrite the equation in 2 cases.

Case 1: $i=1,2,3$

$$ \lambda\frac{\partial}{\partial \epsilon_{j}}(\epsilon_1 + \epsilon_2 + \epsilon_3) + 2\mu\frac{\partial}{\partial \epsilon_{j}}\epsilon_{i} \\[2.5px] =\lambda\frac{\partial}{\partial \epsilon_{j}}(\epsilon_1 + \epsilon_2 + \epsilon_3) + 2\mu\delta_{ij}$$

Can someone comment how the first part simplifies really? Final version is given in the answer, I am trying to reach there.

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You switched incorrectly from $\lambda\,\mathrm{tr}(\epsilon)$ to $\lambda\epsilon_i$; the corresponding term should be $\lambda(\epsilon_1+\epsilon_2+\epsilon_3)$ for $i=1,2,3$ and $0$ otherwise. This would give you

$$\mathbb{C}_{ij}=\begin{cases}\lambda+2\mu&i=j=1,2,3\\ \lambda&i=1,2,3\neq j\\ 2\mu&i=j=4,5,6\\0&\mathrm{otherwise} \end{cases}$$

See p. 62 here, for example.

EDIT: The question was edited to ask about clarification for this derivation. Generalized Hooke's Law can be expressed conveniently in nonreduced notation $$\sigma_{ij}=\lambda\epsilon_{kk}\delta_{ij}+2\mu\epsilon_{ij}$$ but less conveniently in Voight or reduced notation:$$\sigma_{i}=\begin{cases}\lambda(\epsilon_1+\epsilon_2+\epsilon_3)+2\mu\epsilon_{i}&i=1,2,3\\2\mu\epsilon_i&\mathrm{otherwise}\end{cases}$$

So we have $$\mathbb{C}_{ij}=\frac{\partial \sigma_i}{\partial \epsilon_j}$$ $$\mathbb{C}_{ij}=\begin{cases}\frac{\partial}{\partial \epsilon_j}[\lambda(\epsilon_1+\epsilon_2+\epsilon_3)+2\mu\epsilon_{i}]&i=1,2,3\\\frac{\partial}{\partial \epsilon_j}(2\mu\epsilon_i)&\mathrm{otherwise}\end{cases}$$

In the first case, the $\frac{\partial}{\partial \epsilon_j}[\lambda(\epsilon_1+\epsilon_2+\epsilon_3)]$ term always picks out the single $\epsilon$ component corresponding whatever $j$ is, thus always giving $\lambda$, whereas the second term is $2\mu$ only if $i=j$ and $0$ otherwise. In the second case, the term is $2\mu$ only if $i=j$ and $0$ otherwise. We thus obtain our four cases. Does this clarify things at all?

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