# Diff of /branches/stage3.0/doc/user/heatedblock.tex

revision 2583 by jfenwick, Fri Jul 31 05:39:36 2009 UTC revision 2584 by jfenwick, Wed Aug 5 02:44:51 2009 UTC
# Line 27  where the stress $\sigma$ is given by Line 27  where the stress $\sigma$ is given by
27  \end{eqnarray}  \end{eqnarray}
28  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
29  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
32  comparison to the \eqn{WAVE stress}  comparison to the \eqn{WAVE stress}
33  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
# Line 35  to bring in stress due to volume changes Line 35  to bring in stress due to volume changes
35
36  Our domain is the unit cube  Our domain is the unit cube
37  \begin{eqnarray} \label{HEATEDBLOCK natural location}  \begin{eqnarray} \label{HEATEDBLOCK natural location}
38  \Omega=\{(x\hackscore{i} | 0 \le x\hackscore{i} \le 1 \}  \Omega=\{(x\hackscore{i}) | 0 \le x\hackscore{i} \le 1 \}
39  \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
41  \begin{eqnarray} \label{HEATEDBLOCK natural}  \begin{eqnarray} \label{HEATEDBLOCK natural}
# Line 89  q\hackscore{i}(x)=\left\{ Line 89  q\hackscore{i}(x)=\left\{
89  \right.  \right.
90
91  Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$  Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$
92  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,
93  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
94  in~\eqn{HEATEDBLOCK linear elastic} to zero.  in~\eqn{HEATEDBLOCK linear elastic} to zero.
95

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