## 2 Coupled Thermo-Mechanical Phase-Field Modeling
### Energy Imbalance  
According to the rules of thermodynamics, i.e., the Clausius-Duhem inequality, the local dissipation energy of power produced by macroscopic and microscopic forces is considered for this thermodynamic coupling problem [35]:  
\[D=\boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}+K\dot{d}+\mathbf{H} \cdot\
abla\dot{d}-\rho(\dot{\psi}+\dot{T}\eta)-\frac{1}{T}\
abla T\cdot \mathbf{q}\geq 0 \tag{1}\]  
where \(\boldsymbol{\sigma}\) and \(\varepsilon\) denote the Cauchy stress tensor and strain, respectively; \(\psi\) and \(\eta\) represent the Helmholtz free energy and entropy; \(K\) is the internal micro-force; \(\rho\) represents the density; and \(\mathbf{H}\) is the micro-traction on the crack surface.  
Elastic energy \(\psi_{e}\), fracture energy \(\psi_{c}\), and thermal energy \(\psi_{T}\) make up the Helmholtz free energy.  
\[\psi=\psi(\varepsilon,d,\
abla d,T)=\psi_{e}(\varepsilon,d,T)+\psi_{c}(d, \
abla d,T)+\psi_{T}(T) \tag{2}\]  
For the above free energy equation, the rate of change of free energy is as follows.  
\[\dot{\psi}=\dot{\psi}_{e}+\dot{\psi}_{c}+\dot{\psi}_{T} \tag{3}\]  
\[\left\{\begin{array}{l}\dot{\psi}_{e}=\frac{\partial\psi_{e}}{\partial t}: \dot{\varepsilon}+\frac{\partial\psi_{c}}{\partial d}\dot{d}+\frac{\partial \psi_{c}}{\partial T}\dot{T}\\ \dot{\psi}_{c}=\frac{\partial\psi_{c}}{\partial t}\dot{d}+\frac{\partial\psi_ {c}}{\partial(\
abla d)}\cdot\
abla d+\frac{\partial\psi_{c}}{\partial T}\dot{ T}\\ \dot{\psi}_{T}=\frac{\partial\psi_{c}}{\partial T}\dot{T}\end{array}\right. \tag{4}\]  
We can obtain the below equation after substituting Equation (4) into Equation (3).  
\[\dot{\psi}=\frac{\partial\psi_{e}}{\partial\varepsilon}:\dot{\varepsilon}+ \left(\frac{\partial\psi_{e}}{\partial d}+\frac{\partial\psi_{c}}{\partial d }\right)\dot{d}+\frac{\partial\psi_{c}}{\partial(\
abla d)}\cdot\
abla\dot{d }+(\frac{\partial\psi_{e}}{\partial T}+\frac{\partial\psi_{c}}{\partial T}) \dot{T} \tag{5}\]  
It can be rewritten as follows by taking Equation (5) into the Clausius-Duhem Inequality (1).  
\[\left(\sigma-\rho\frac{\partial\psi}{\partial\varepsilon}\right):\dot{ \varepsilon}+\left(K-\rho\frac{\partial\psi}{\partial d}\right)\dot{d}+\left( \mathbf{H}-\rho\frac{\partial\psi}{\partial(\
abla d)}\right)\cdot\
abla d- \left(\rho\eta+\rho\frac{\partial\psi}{\partial T}\right)\dot{T}-\frac{1}{T} \
abla T\cdot\mathbf{q}\geq 0 \tag{6}\]  
It is important to note Inequality (6) needs to meet the thermodynamic processes in any case. As a result, the coefficients of the dissipative terms are not negative, whereas the coefficients of the non-dissipative terms must be zero. According to [PERSON] [41], the thermoelastic constitutive relation is depicted below.  
\[\left\{\begin{array}{l}\boldsymbol{\sigma}=\rho\frac{\partial\psi}{\partial t }=\rho\frac{\partial\psi_{c}}{\partial\varepsilon}\\ \mathbf{H}=\rho\frac{\partial\psi}{\partial(\
abla d)}=\rho\frac{\partial\psi _{c}}{\partial(\
abla d)}\\ K=\rho\frac{\partial\psi}{\partial d}=\rho\frac{\partial\psi_{c}}{\partial d}+ \rho\frac{\partial\psi_{c}}{\partial d}\\ \eta=-\frac{\partial\psi}{\partial T}\end{array}\right. \tag{7}\]  
The rest of the dissipation inequality is also known as the heat conduction inequality.  
\[-\
abla T\cdot\mathbf{q}\geq 0 \tag{8}\]