13th International Conference on Fracture June 16–21, 2013, Beijing, China -3- ( ) , 0 , 2 , 0 T x t x a t x ∂ = = > ∂ (3) ( ) 0 , , 0,0 2 T x t T t x a = = < < (4) where k(T), ρ(T) and Cp(T) are the temperature-dependent thermal conductivity, mass density and specific heat capacity, respectively. And t represents time. Obviously this is a transient nonlinear heat conduction problem. In order to solve Eq. (1), we linearize this partial differential heat conduction equation using the Kirchhoff transformation [20], namely ( ) ( ) 0 0 1 T T T k T dT k θ = ∫ (5) where k0 is the thermal conductivity at reference temperature T0. Under the transformation (5), equations (1), (2), (3) and (4) reformulate as ( ) ( ) 2 2 , , 1 , 0 2 , 0 x t x t x a t x t θ θ κ ∂ ∂ = < < > ∂ ∂ (6) ( ) ( ) ( ) 0 , , , 0, 0 x t k h x t x t x θ θ θ∞ ∂ = − = > ∂ (7) ( ) , 0 , 2 , 0 x t x a t x θ∂ = = > ∂ (8) ( ) i , , 0 2 , 0 x t x a t θ θ = < < = (9) However, in Eq. (6) ( ) ( ) ( ) p k T T C T κ ρ = (10) is still temperature-dependent. For the convenience of derivation and calculation, we determine a 0κ with the use of Eq. (11) to replace in Eq. (6).Therefore, Eq.(6) becomes a total linear equation. ( ) ( ) ( ) 0 0 0 p 1 T T k T dT T T T C T κ ρ ∞ ∞ = − ∫ (11) Applying the variable separation approach, the expression of ( ) ,x t θ is obtained. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 0 2 2 i 2 2 1 sin , 2 cos m m m t m m m m H L x t L x e H H L κ β β β θ θ θ θ β β β ∞ − ∞ ∞ = + = + − − + + ∑ (12) in which 0 h k H= , L=2a, and mβ is determined by the following Eq. (13). Therefore, the temperature fields T(x,t) can be acquired from ( ) ,x t θ in Eq. (12) making the use of their relationship in Eq. (5).
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