27 
\end{eqnarray} 
\end{eqnarray} 
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In this formula $\lambda$ and $\mu$ are the Lame coefficients, $\alpha$ is the 
In this formula $\lambda$ and $\mu$ are the Lame coefficients, $\alpha$ is the 
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temperature expansion coefficient, $T$ is the temperature distribution and $T_{ref}$ a reference temperature. Note that 
temperature expansion coefficient, $T$ is the temperature distribution and $T_{ref}$ a reference temperature. Note that 
30 
\eqn{HEATEDBLOCK general problem} is similar to eqn{WAVE general problem} introduced in section~\Sec{WAVE CHAP} but the 
\eqn{HEATEDBLOCK general problem} is similar to \eqn{WAVE general problem} introduced in section~\Sec{WAVE CHAP} but the 
31 
inertia term $\rho u\hackscore{i,tt}$ has been dropped as we assume a static scenario here. Moreover, in 
inertia term $\rho u\hackscore{i,tt}$ has been dropped as we assume a static scenario here. Moreover, in 
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comparison to the \eqn{WAVE stress} 
comparison to the \eqn{WAVE stress} 
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definition of stress $\sigma$ in \eqn{HEATEDBLOCK linear elastic} an extra term is introduced 
definition of stress $\sigma$ in \eqn{HEATEDBLOCK linear elastic} an extra term is introduced 
35 


36 
Our domain is the unit cube 
Our domain is the unit cube 
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\begin{eqnarray} \label{HEATEDBLOCK natural location} 
\begin{eqnarray} \label{HEATEDBLOCK natural location} 
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\Omega=\{(x\hackscore{i}  0 \le x\hackscore{i} \le 1 \} 
\Omega=\{(x\hackscore{i})  0 \le x\hackscore{i} \le 1 \} 
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\end{eqnarray} 
\end{eqnarray} 
40 
On the boundary the normal stress component is set to zero 
On the boundary the normal stress component is set to zero 
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\begin{eqnarray} \label{HEATEDBLOCK natural} 
\begin{eqnarray} \label{HEATEDBLOCK natural} 
89 
\right. 
\right. 
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\end{equation} 
\end{equation} 
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Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$ 
Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$ 
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are constant we may use $Y\hackscore{i}=\lambda+\frac{2}{3} \mu) \; \alpha \; T\hackscore{i}$. However, 
are constant we may use $Y\hackscore{i}=(\lambda+\frac{2}{3} \mu) \; \alpha \; T\hackscore{i}$. However, 
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this choice would lead to a different natural boundary condition which does not set the normal stress component as defined 
this choice would lead to a different natural boundary condition which does not set the normal stress component as defined 
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in~\eqn{HEATEDBLOCK linear elastic} to zero. 
in~\eqn{HEATEDBLOCK linear elastic} to zero. 
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