## 2 Fundamentals of the Continuous-Discontinuous Thermo-Mechanical Coupling Model  
### The Continuous-Discontinuous Heat Conduction Model  
As shown in Figure 1, the entire continuous-discontinuous medium is divided into triangular elements. The adjacent triangular elements on both sides of the crack do not share nodes, while the remaining adjacent triangular elements share nodes. The temperature distribution within a triangular element is represented by linear interpolation of the temperature at the three nodes of the triangular element. In addition, heat conduction inside the triangular element and heat transfer between adjacent triangular elements may occur through the crack. Therefore, we can obtain the temperature distribution of the entire continuum region. Figure 1: (**a**) continuous–discontinuous medium, (**b**) continuous–discontinuous medium calculation based on the initial mesh. For any given mass \(M\), the temperature change is expressed as follows:  
\[\frac{\partial T}{\partial t}=\frac{Q_{total}}{C_{p}M} \tag{1}\]  
where \(Q_{total}\) is the net heat flux flowing into the mass \(M\) per unit time, \(C_{p}\) is the specific heat capacity, and \(M\) is the mass. Figure 1a shows that the continuous-discontinuous medium is divided into finite element meshes of triangular elements. Then, the nodes at the cracks are separated, as shown in Figure 1b. The entire continuous-discontinuous medium is discretized into triangular elements composed of nodes. The temperature distribution of the entire continuous-discontinuous medium can be roughly estimated from the temperature at the nodes of these triangle elements. The temperature distribution in each triangular element can be obtained by linear interpolation based on the element's three nodes. The mass of the triangle element is distributed evenly among the three nodes of the triangular element. The mass of each node is equal to 1/3 of the mass of all the triangular elements that share the node. Based on the relationship between the elements and nodes illustrated in Figure 1b, we will explain how to calculate the temperature field for the entire continuous-discontinuous medium. Taking node 4 in Figure 1b as an example, the triangular elements sharing node 4 are \(\Delta\)345, \(\Delta\)456, and \(\Delta\)134. Since the temperature of node 4 may be different from the temperature of nodes 1, 3, 5, and 6, heat conduction may occur in these triangular elements that share node 4. Take the triangular element \(\Delta\)345 as an example. We assume that the temperature field distribution in the triangular element follows a linear distribution. The temperature gradient at any point in the same triangular element is constant, which can be expressed as:  
\[\frac{\partial T}{\partial x_{i}}=\frac{1}{A}\int\limits_{A}\frac{\partial T }{\partial x_{i}}dA=\frac{1}{A}\int\limits_{s}Tn_{i}ds=\frac{1}{A}\sum\limits _{m=1}^{3}\overline{T}^{m}\in_{ij}\Delta x_{j}^{m} \tag{2}\]  
where \(A\) is the area of the triangular element, \(n_{i}\) is the outer normal vector, \(\overline{T}^{m}\) is the average temperature of the edge \(m\), \(\Delta x_{j}^{m}\) is the difference between the coordinate components of the two vertices of the edge \(m\) side, and \(\in_{ij}\) is the two-dimensional permutation tensor \(\in= \begin{pmatrix}0&1\\ -1&0\end{pmatrix} \). Based on [PERSON]'s law of heat conduction, the heat flow rate per unit cross-sectional area in the \(i\) direction is given by:  
\[q_{i}=-k_{ij}\frac{\partial T}{\partial x_{j}} \tag{3}\]  
where \(k_{ij}\) is the thermal conductivity tensor, \(T\) is the temperature, and \(x_{j}\) is the position vector. The temperature of node 4 in the next time step \(\Delta t\) is as follows:  
\[T_{4}^{t+\Delta t}=T_{4}^{t}+\frac{Q_{total\to 4}}{C_{p}M}\Delta t \tag{4}\]  
The temperature of the other nodes can be obtained similarly according to the above step. Based on this, we can obtain the evolution of the temperature field in the entire solution domain.