Actions during hardening 89
As the two halves are fully connected to each other, these free thermal
strains cannot occur. Due to compatibility of strains, an overall strain ε
will occur (Figure 4.12c), now leading to thermal stresses σ 1 and σ 2 in the
two halves, which can be calculated as follows (compressive stresses taken
as positive):
σ 1
1
=
−
E(
)
ε ε
(4.5)
σ 2
2
=
−
E(
)
ε ε
(4.6)
By expressing equilibrium (σ 1 + σ 2 = 0), it can be shown that:
ε
ε ε
θ
θ α
=
+ =
+
⋅
1
2
1
2
2
2
∆
∆
t
(4.7)
and
σ
σ
θ
θ
α
1
2
1
2
2
= − =
−
⋅
∆
∆
E t
⋅
(4.8)
Suppose that at time tʹ the imposed temperature variations are withdrawn (Figure  4.12d). The final stress and strain conditions can then be
obtained by:
ε
ε ε
θ
θ α
θ
θ α
b
t
t
= + ′ =
+
⋅ +
−
−
⋅ =
∆
∆
∆
∆
1
2
1
2
2
2
0
(4.9)
σ
σ
σ σ
θ
θ α
θ
θ α
1
2
1
1
1
2
1
2
2
2
0
b
b
t
t
E
E
=
= + ′ =
−
+
−
+
=
∆
∆
∆
∆
(4.10)
In other words, no remaining stresses or strains are obtained when the two
halves have returned to their original temperatures.
Now consider the more complicated case of a time and location dependent Young’s modulus, as illustrated in Figure 4.13. It can be shown that
the remaining strain and stresses after returning to the original temperature are obtained by:
ε
ε ε
θ
θ
θ
θ
b = + ′ =
+
+
−
′
+ ′
′ + ′
E
E
E E
E
E
E
1
2
1
2
1
2
1
∆
∆
∆
∆
1
2
1
2
E E 2





⋅ α t
(4.11)
σ
σ
θ
1
2
1 2
1
2
1 2
1
2
1
b
b
E E
E E
E E
E E
= −
=
+
−
′ ′
′ + ′





⋅
−
∆
∆ ∆θ α
2
2
⋅ t
(4.12)