8.3 Formulation of the Problem
307
deformation (trace of the deformation tensor) is the only source of heat production
and hence there is a lack of curvature tensor in Eq. (8.21). In the case of plane stress
state (8.8), the trace of deformation tensor ε ii can be presented in the following form:
ε ii = (1 − 2ν)
u , x +
1
2
w , x
2 + zψ , x
+ 2 (1 + ν) α θ.
(8.22)
Substituting (8.22) into (8.21) and assuming that heat gradients along the O Zaxis are essentially larger than the gradient along the remaining O X- and OY -axes,
Eq. (8.21) takes the following form
∂
2
θ
∂ z 2 =
ρ c v
λ
+
2 (1 + ν) Eα
2 T 0
λ (1 − 2ν)
∂ θ
∂ t
+
+
Eα T 0
λ
∂
2 u
∂ t ∂ x
+
∂ w
∂ x
∂
2 w
∂ t ∂ x
+ z
∂
2
ψ
∂ t ∂ x
.
(8.23)
Let us introduce the coefficients of heat diffusion D = λ/ρ c v and relaxation
R = Eα
2 T 0 /ρ c v . Then, Eq. (8.23) can be presented as follows:
(1 + )
∂ θ
∂ t
= D
∂
2
θ
∂ z 2 −
R
α
∂
2 u
∂ t ∂ x
+
∂ w
∂ x
∂
2 w
∂ t ∂ x
+ z
∂
2
ψ
∂ t ∂ x
,
(8.24)
where = 2R (1 + ν)/(1 − 2ν) .
In the case of plane deformation, the trace of the deformation tensor is yielded by
relations (8.9) and takes the following form:
ε ii =
1 − 2ν
1 − ν
u , x +
1
2
w , x
2 + zψ , x
+
1 + ν
1 − ν
α θ.
(8.25)
Substituting (8.25) into (8.21) and taking into account heat gradients only along
the axis, the following equation is derived:
∂
2
θ
∂ z 2 =
ρ c v
λ
+
(1 + ν)
(1 − ν) (1 − 2ν)
Eα
2 T 0
λ
∂ θ
∂ t
+
+
Eα T 0
λ (1 − ν)
∂
2 u
∂ t ∂ x
+
∂ w
∂ x
∂
2 w
∂ t ∂ x
+ z
∂
2
ψ
∂ t ∂ x
,
(8.26)
which can be rewritten in terms of heat diffusion and relaxation:
1 + ˜
∂ θ
∂ t
= D
∂
2
θ
∂ z 2 −
˜
R
α
∂
2 u
∂ t ∂ x
+
∂ w
∂ x
∂
2 w
∂ t ∂ x
+ z
∂
2
ψ
∂ t ∂ x
,
(8.27)
where ˜
R = R/(1 − ν) , ˜
= ˜
R (1 + ν)/(1 − 2ν) .