Chapter 9, Thermal Solid Mechanics Video Solutions, A First Course in Continuum Mechanics

Problem 1

Consider a more general constitutive model for a thermoelastic solid of the form
$$
\begin{aligned}
&\boldsymbol{P}=\widehat{\boldsymbol{P}}(\boldsymbol{F}, \Theta, \boldsymbol{g}), \quad \boldsymbol{Q}=\widehat{\boldsymbol{Q}}(\boldsymbol{F}, \Theta, \boldsymbol{g}) \\
&\eta_{m}=\widehat{\eta}_{m}(\boldsymbol{F}, \Theta, \boldsymbol{g}), \quad \Psi=\hat{\Psi}(\boldsymbol{F}, \Theta, \boldsymbol{g})
\end{aligned}
$$
where $\boldsymbol{F}=\nabla^{x} \varphi$ and $\boldsymbol{g}=\nabla^{x} \Theta .$ Show that such a model satisfies the Clausius-Duhem inequality if and only if all the following are true:
(i) $\widehat{\boldsymbol{P}}, \hat{\eta}_{m}$ and $\hat{\Psi}$ are independent of $\boldsymbol{g}$
(ii) $\widehat{\boldsymbol{P}}(\boldsymbol{F}, \Theta)=\rho_{0} D_{\boldsymbol{F}} \hat{\Psi}(\boldsymbol{F}, \Theta)$ and $\hat{\eta}_{m}(\boldsymbol{F}, \Theta)=-D_{\Theta} \hat{\Psi}(\boldsymbol{F}, \Theta)$
(iii) $\widehat{\boldsymbol{Q}}(\boldsymbol{F}, \Theta, \boldsymbol{g}) \cdot \boldsymbol{g} \leq 0$ for all admissible $(\boldsymbol{F}, \Theta, \boldsymbol{g})$.

Ameer Said

Numerade Educator

03:00

Problem 2

Consider a thermoelastic body with material and Fourier-Stokes heat flux vector fields given by
$$
\boldsymbol{Q}=\widehat{\boldsymbol{Q}}(\boldsymbol{F}, \Theta, \boldsymbol{g}), \quad \boldsymbol{q}_{m}=\widehat{\boldsymbol{q}}_{m}(\boldsymbol{F}, \Theta, \boldsymbol{g})
$$
where $\boldsymbol{g}=\nabla^{x} \Theta .$ By definition, such a body is thermally isotropic if
$$
\widehat{\boldsymbol{q}}_{m}\left(\boldsymbol{F} \boldsymbol{A}, \Theta, \boldsymbol{\Lambda}^{T} \boldsymbol{g}\right)=\widehat{\boldsymbol{q}}_{m}(\boldsymbol{F}, \Theta, \boldsymbol{g})
$$
for all rotation tensors $\boldsymbol{\Lambda}$ and all $(\boldsymbol{F}, \Theta, \boldsymbol{g})$ with det $\boldsymbol{F}>0$ and $\Theta>0$
(a) Assuming $\widehat{\boldsymbol{Q}}(\boldsymbol{F}, \Theta, \boldsymbol{g})=-\overline{\boldsymbol{K}}(\boldsymbol{C}, \Theta) \boldsymbol{g}$, where $\boldsymbol{C}=\boldsymbol{F}^{T} \boldsymbol{F}$, find an expression for $\widehat{\boldsymbol{q}}_{m}(\boldsymbol{F}, \Theta, \boldsymbol{g})$
(b) Show that a thermoelastic solid is thermally isotropic if and only if the thermal conductivity function $\bar{K}$ takes the form
$$
\overline{\boldsymbol{K}}(\boldsymbol{C}, \Theta)=\kappa_{0}\left(\mathcal{I}_{C}, \Theta\right) \boldsymbol{I}+\kappa_{1}\left(\mathcal{I}_{C}, \Theta\right) \boldsymbol{C}+\kappa_{2}\left(\mathcal{I}_{C}, \Theta\right) \boldsymbol{C}^{-1}
$$
for some scalar-valued functions $\kappa_{0}, \kappa_{1}$ and $\kappa_{2}$.

Joseph Liao

Numerade Educator

01:56

Problem 3

Consider a thermoelastic body with response functions satisfying the conditions of Results $9.1$ and $9.3$. Suppose the body has a reference configuration with uniform mass density $\rho_{*}$ and uniform temperature $\Theta_{*} .$ Moreover, suppose the free energy response function is of the form
$$
\bar{\Psi}(\boldsymbol{C}, \Theta)=\frac{1}{2} \boldsymbol{G}: \mathbf{J}(\boldsymbol{G})+\vartheta \boldsymbol{M}: \boldsymbol{G}-\frac{1}{2} c \vartheta^{2}
$$
where $\mathbf{J}$ is a constant fourth-order tensor with major and minor symmetry (both right and left), $\boldsymbol{M}$ is a constant, symmetric second-order tensor, $c>0$ is a constant scalar, $\boldsymbol{G}=\frac{1}{2}(\boldsymbol{C}-\boldsymbol{I})$ is the Green-St. Venant strain tensor and $\vartheta=\Theta-\Theta_{*}$ is the temperature deviation.
(a) Show that the response functions $\widehat{\boldsymbol{P}}$ and $\hat{\eta}_{m}$ corresponding to $\bar{\Psi}$ are
$$
\begin{aligned}
\widehat{\boldsymbol{P}}(\boldsymbol{F}, \Theta) &=\rho_{*} \boldsymbol{F}[\mathbf{J}(\boldsymbol{G})+\vartheta \boldsymbol{M}] \\
\widehat{\eta}_{m}(\boldsymbol{F}, \Theta) &=c \vartheta-M: G
\end{aligned}
$$
Moreover, verify that the reference configuration is stress-free.
(b) Show that the elasticity tensor $\mathbf{A}_{*}$, thermal stress tensor $\boldsymbol{D}_{*}$ and entropy parameter $\beta_{*}$ corresponding to $\bar{\Psi}$ are given by
$$
\mathbf{A}_{*}=\rho_{*} \mathbf{J}, \quad \boldsymbol{D}_{*}=\rho_{*} \boldsymbol{M}, \quad \beta_{*}=c
$$

Narayan Hari

Numerade Educator

02:16

Problem 4

Consider a stress response function of the form $\widehat{\boldsymbol{P}}(\boldsymbol{F}, \Theta)=$ $\rho_{*} D_{F} \widehat{\Psi}(\boldsymbol{F}, \Theta)$, where $\widehat{\Psi}(\boldsymbol{F}, \Theta)=\bar{\Psi}(\boldsymbol{C}, \Theta) .$ Assuming a stressfree reference state, show that the elasticity tensor $\mathbf{A}_{*}$ and the thermal stress tensor $\boldsymbol{D}_{*}$ satisfy
$$
\mathrm{A}_{i j k l}^{*}=\rho_{*} \frac{\partial^{2} \widehat{\Psi}}{\partial F_{i j} \partial F_{k l}}\left(\boldsymbol{I}, \Theta_{*}\right)=4 \rho_{*} \frac{\partial^{2} \bar{\Psi}}{\partial C_{i j} \partial C_{k l}}\left(\boldsymbol{I}, \Theta_{*}\right)
$$
and
$$
D_{i j}^{*}=\rho_{*} \frac{\partial^{2} \widehat{\Psi}}{\partial \Theta \partial F_{i j}}\left(\boldsymbol{I}, \Theta_{*}\right)=2 \rho_{*} \frac{\partial^{2} \bar{\Psi}}{\partial \Theta \partial C_{i j}}\left(\boldsymbol{I}, \Theta_{*}\right)
$$
Deduce that $\mathbf{A}_{*}$ necessarily has major and minor symmetry (both right and left) and that $\boldsymbol{D}_{*}$ is necessarily symmetric.

Chai Santi

Numerade Educator

01:19

Problem 5

Assuming no coupling $\left(\boldsymbol{D}_{*}=\boldsymbol{O}\right)$, consider the linearized thermoelastodynamics equation for the temperature disturbance in
a thermoelastic body with reference configuration $B=\{\boldsymbol{X} \in$ $\left.\mathbb{E}^{3} \mid 0<X_{i}<1\right\}$, namely
$$
\begin{array}{lll}
\rho_{*} \alpha_{*} \Theta^{(1)}=\nabla^{x} \cdot\left[\boldsymbol{K}_{*} \nabla^{x} \Theta^{(1)}\right]+\rho_{*} R^{(1)} & \text { in } \quad B \times[0, T] \\
\Theta^{(1)}=\xi^{(1)} & \text { in } \quad \Gamma_{\theta} \times[0, T] \\
\boldsymbol{K}_{*} \nabla^{x} \Theta^{(1)} \cdot \boldsymbol{N}=-\zeta^{(1)} & \text { in } \quad \Gamma_{q} \times[0, T] \\
\Theta^{(1)}(\cdot, 0)=\Theta_{0}^{(1)} & \text { in } B .
\end{array}
$$
Suppose that there is no heat supply, so $R^{(1)}=0$, that the body is thermally isotropic, so $\boldsymbol{K}_{*}=\kappa_{*} \boldsymbol{I}$, and that the temperature disturbance vanishes on the entire boundary, so $\xi^{(1)}=0, \Gamma_{\theta}=$ $\partial B$, and $\Gamma_{q}=\emptyset$. Moreover, assume that the initial condition can be expanded in a Fourier sine series as
$$
\Theta_{0}^{(1)}(\boldsymbol{X})=\sum_{k_{1}, k_{2}, k_{3}=1}^{\infty} A_{k_{1} k_{2} k_{3}} \sin \left(k_{1} \pi X_{1}\right) \sin \left(k_{2} \pi X_{2}\right) \sin \left(k_{3} \pi X_{3}\right)
$$
where $A_{k_{1} k_{2} k_{5}}$ are scalar constants. Use separation of variables to find the temperature disturbance field $\Theta^{(1)}(\boldsymbol{X}, t)$ for all $\boldsymbol{X} \in$ $B$ and $t \geq 0$

Manik Pulyani

Numerade Educator

01:47

Problem 6

Show that the linearized balance of momentum and energy equations for a mechanically and thermally isotropic body take the forms
$$
\begin{gathered}
\rho_{*} \ddot{\boldsymbol{u}}^{(1)}=\mu_{*} \Delta^{x} \boldsymbol{u}^{(1)}+\left(\lambda_{*}+\mu_{*}\right) \nabla^{x}\left(\nabla^{x} \cdot \boldsymbol{u}^{(1)}\right) \\
+m_{*} \nabla^{x} \Theta^{(1)}+\rho_{*} \boldsymbol{b}_{m}^{(1)} \\
\rho_{*} \alpha_{*} \Theta^{(1)}=\kappa_{*} \Delta^{x} \Theta^{(1)}+\Theta_{*} m_{*} \nabla^{x} \cdot \dot{\boldsymbol{u}}^{(1)}+\rho_{*} R^{(1)}
\end{gathered}
$$
where $\rho_{*}$ is the mass density, $\Theta_{*}$ is the reference temperature, $\alpha_{*}$ is the specific heat, $\lambda_{*}$ and $\mu_{*}$ are the Lamé constants, $m_{*}$ is the thermal stress coefficient and $\kappa_{*}$ is the thermal conductivity.

Ajay Singhal

Numerade Educator

01:35

Problem 7

Assuming no body force and no heat supply, consider the equations for a fully isotropic body derived in Exercise 6 , namely
$$
\begin{aligned}
&\ddot{\boldsymbol{u}}^{(1)}=\mu \Delta^{x} \boldsymbol{u}^{(1)}+(\lambda+\mu) \nabla^{x}\left(\nabla^{x} \cdot \boldsymbol{u}^{(1)}\right)+m \nabla^{x} \Theta^{(1)} \\
&\dot{\Theta}^{(1)}=\kappa \Delta^{x} \Theta^{(1)}+\nu \nabla^{x} \cdot \dot{\boldsymbol{u}}^{(1)}
\end{aligned}
$$
where $\lambda=\lambda_{*} / \rho_{*}, \mu=\mu_{*} / \rho_{*}, m=m_{*} / \rho_{*}, \kappa=\kappa_{*} /\left(\rho_{*} \alpha_{*}\right)$ and $\nu=\Theta_{*} m_{*} /\left(\rho_{*} \alpha_{*}\right) .$ We suppose $m \neq 0$ and $\nu \neq 0$ so that the equations are fully coupled. Here we study plane progressive

Narayan Hari

Numerade Educator

02:12

Problem 8

Consider the stress-strain relation for a linearized, isotropic, thermoelastic solid with Lamé constants $\lambda_{*}, \mu_{*}$ and thermal stress coefficient $m_{*}$
$$
\boldsymbol{P}^{(1)}=\lambda_{*}\left(\operatorname{tr} \boldsymbol{E}^{(1)}\right) \boldsymbol{I}+2 \mu_{*} \boldsymbol{E}^{(1)}+m_{*} \Theta^{(1)} \boldsymbol{I}
$$
Show that this relation can be inverted as
$$
\boldsymbol{E}^{(1)}=a_{*}\left(\operatorname{tr} \boldsymbol{P}^{(1)}\right) \boldsymbol{I}+2 b_{*} \boldsymbol{P}^{(1)}+c_{*} \Theta^{(1)} \boldsymbol{I}
$$
where
$$
a_{*}=-\frac{\lambda_{*}}{2 \mu_{*}\left(3 \lambda_{*}+2 \mu_{*}\right)}, \quad b_{*}=\frac{1}{4 \mu_{*}}, \quad c_{*}=-\frac{m_{*}}{3 \lambda_{*}+2 \mu_{*}}
$$
Remark: The above result shows that, when the elasticity tensor $\mathbf{A}_{*}$ is isotropic and the thermal stress tensor $\boldsymbol{D}_{*}$ is spherical, so are the compliance tensor $\mathbf{G}_{*}$ and the thermal strain tensor $\boldsymbol{H}_{*}$ In this case, the thermal stress in the absence of strain is a pressure, whereas the thermal strain in the absence of stress is a dilatation. The constant $c_{*}$ is typically called the coefficient of thermal expansion.

Narayan Hari

Numerade Educator