into the volume element; second, it changes because there is an entropy source due to
irreversible phenomena inside the volume element. The entropy source is always a
nonnegative quantity, since entropy can only be created, never destroyed, unlike
energy.
Entropy is the measure of how much energy is unavailable for work. Imagine an isolated and
closed system with some hot objects and some cold objects. Work can be done as heat is
transferred from the hot to the cooler objects; however, once this transfer has occurred, it is
impossible to extract additional work from them alone. Energy is always conserved, but
Table 5.1 Return mapping algorithm classical theory rate-dependent model
Update strain
ε n + 1 = ε n + — Δμ
Compute trial state
S
tr
nþ1 = S n þ 1 2 Φ n
ð
Þ 2μΔe nþ1
Compute trial yield function F
tr
nþ1 = S
tr
nþ1 2 X
D
n
2 1 2 Φ
ð
Þ
ffiffi
2
3
q
K α n
ð Þ
IF F
tr
nþ1 > 0THEN
Call Newton local and solve g(Δγ) = 0 for Δγ
g Δγ
ð Þ ¼ S n À X
Φ
n
2 þ 1 À Φ
ð
Þ 2μΔe nþ1 þ b nþ1 ΔγX
Φ
n
2 þ
n
2 S n À X
Φ
n
À
Á : 1 À Φ
ð
Þ2μΔe nþ1 þ b nþ1 ΔγX
Φ
n
Â
Ã É 1=2
À 1 À Φ
ð
Þ
ffiffi ffi
2
3
r
K α n þ
ffiffi ffi
2
3
r
Δγ
!
À Δγ 1 À Φ
ð
Þ 2μ þ a nþ1
½
À Θ
Δγη
Δt
where a nþ1 =
c1 1 2 Φ
ð
Þ
1þc 0
2
1 2 Φ
ð
Þ1 2 β
ð
Þ Δγ , b nþ1 =
c2
c1 a nþ1
Update
n nþ1
Snþ1 2 X
Φ
nþ1
Snþ1 2 X
Φ
nþ1
k
k
=
Bn
Bn
k k where B n = S
tr
nþ1 2 X
Φ
n þ b nþ1 ΔγX
Φ
n
α nþ1 = α n þ
ffiffi
2
3
q
Δγ
ε
vp
nþ1 = ε
vp
n þ Δγ
Bn
Bn
k k
X
Φ
nþ1 = X
Φ
n þ a nþ1 Δγ
Snþ1 2 X
Φ
nþ1
Snþ1 2 X
Φ
nþ1
k
k
2
c
0
2
c1 X
Φ
n
ξ
Φ
nþ1 1 2 Φ
ð
ÞK α nþ1
ð
Þ
Bn
Bn
k k
S nþ1 = ξ
Φ
nþ1 þ X
Φ
nþ1
σ nþ1 = ҡ 1 2 Φ
ð
Þtr ε nþ1
ð
Þ b I þ 2μ 1 2 Φ
ð
Þ e nþ1 2 ε
vp
n 2 γ nþ1
Bn
Bn
k k 2 e
θ
nþ1
Compute consistent Jacobian
C
EVPD
nþ1 = 1 2 Φ
ð
Þҡ b I b I þ 2μ 1 2 Φ
ð
Þδ nþ1
Q 2
1
3
b I b I
2 2μ 1 2 Φ
ð
Þθ nþ1
b n nþ1 b n nþ1 2
2μ 1 2 Φ
ð
Þ
Bn
k k Δγ
Q 2 b n nþ1 b n nþ1
ð
Þ :
K4
K3
b n nþ1 X
Φ
n
Where δ nþ1 = 1 2
Δγ2μ 1 2 Φ
ð
Þ
Bn
k k
and θ nþ1 =
1
K3
2
Δγ2μ 1 2 Φ
ð
Þ
Bn
k k
with
K 3 = K 1 þ K 2
K 1 ¼ 1 þ
K
0
3μ þ
anþ1
2μ 1ÀΦ
ð
Þ
K 2 ¼
a
0
nþ1 Δγ
2μ 1ÀΦ
ð
Þ þ
b n nþ1 bnþ1
2μ 1ÀΦ
ð
Þ b nþ1 1 À β
ð
ÞΔγ À 1
½
: X
Φ
n þ
1
2μ 1ÀΦ
ð
Þ
∂Θ
∂Δγ
K 4 = b
0
nþ1 Δγ þ b nþ1
ELSE
Elastic step ð Þ nþ1 = ð Þ
tr
nþ1 (Exit)
END IF
EXIT
214
5 Unified Mechanics of Thermo-mechanical Analysis
irreversible phenomena inside the volume element. The entropy source is always a
nonnegative quantity, since entropy can only be created, never destroyed, unlike
energy.
Entropy is the measure of how much energy is unavailable for work. Imagine an isolated and
closed system with some hot objects and some cold objects. Work can be done as heat is
transferred from the hot to the cooler objects; however, once this transfer has occurred, it is
impossible to extract additional work from them alone. Energy is always conserved, but
Table 5.1 Return mapping algorithm classical theory rate-dependent model
Update strain
ε n + 1 = ε n + — Δμ
Compute trial state
S
tr
nþ1 = S n þ 1 2 Φ n
ð
Þ 2μΔe nþ1
Compute trial yield function F
tr
nþ1 = S
tr
nþ1 2 X
D
n
2 1 2 Φ
ð
Þ
ffiffi
2
3
q
K α n
ð Þ
IF F
tr
nþ1 > 0THEN
Call Newton local and solve g(Δγ) = 0 for Δγ
g Δγ
ð Þ ¼ S n À X
Φ
n
2 þ 1 À Φ
ð
Þ 2μΔe nþ1 þ b nþ1 ΔγX
Φ
n
2 þ
n
2 S n À X
Φ
n
À
Á : 1 À Φ
ð
Þ2μΔe nþ1 þ b nþ1 ΔγX
Φ
n
Â
Ã É 1=2
À 1 À Φ
ð
Þ
ffiffi ffi
2
3
r
K α n þ
ffiffi ffi
2
3
r
Δγ
!
À Δγ 1 À Φ
ð
Þ 2μ þ a nþ1
½
À Θ
Δγη
Δt
where a nþ1 =
c1 1 2 Φ
ð
Þ
1þc 0
2
1 2 Φ
ð
Þ1 2 β
ð
Þ Δγ , b nþ1 =
c2
c1 a nþ1
Update
n nþ1
Snþ1 2 X
Φ
nþ1
Snþ1 2 X
Φ
nþ1
k
k
=
Bn
Bn
k k where B n = S
tr
nþ1 2 X
Φ
n þ b nþ1 ΔγX
Φ
n
α nþ1 = α n þ
ffiffi
2
3
q
Δγ
ε
vp
nþ1 = ε
vp
n þ Δγ
Bn
Bn
k k
X
Φ
nþ1 = X
Φ
n þ a nþ1 Δγ
Snþ1 2 X
Φ
nþ1
Snþ1 2 X
Φ
nþ1
k
k
2
c
0
2
c1 X
Φ
n
ξ
Φ
nþ1 1 2 Φ
ð
ÞK α nþ1
ð
Þ
Bn
Bn
k k
S nþ1 = ξ
Φ
nþ1 þ X
Φ
nþ1
σ nþ1 = ҡ 1 2 Φ
ð
Þtr ε nþ1
ð
Þ b I þ 2μ 1 2 Φ
ð
Þ e nþ1 2 ε
vp
n 2 γ nþ1
Bn
Bn
k k 2 e
θ
nþ1
Compute consistent Jacobian
C
EVPD
nþ1 = 1 2 Φ
ð
Þҡ b I b I þ 2μ 1 2 Φ
ð
Þδ nþ1
Q 2
1
3
b I b I
2 2μ 1 2 Φ
ð
Þθ nþ1
b n nþ1 b n nþ1 2
2μ 1 2 Φ
ð
Þ
Bn
k k Δγ
Q 2 b n nþ1 b n nþ1
ð
Þ :
K4
K3
b n nþ1 X
Φ
n
Where δ nþ1 = 1 2
Δγ2μ 1 2 Φ
ð
Þ
Bn
k k
and θ nþ1 =
1
K3
2
Δγ2μ 1 2 Φ
ð
Þ
Bn
k k
with
K 3 = K 1 þ K 2
K 1 ¼ 1 þ
K
0
3μ þ
anþ1
2μ 1ÀΦ
ð
Þ
K 2 ¼
a
0
nþ1 Δγ
2μ 1ÀΦ
ð
Þ þ
b n nþ1 bnþ1
2μ 1ÀΦ
ð
Þ b nþ1 1 À β
ð
ÞΔγ À 1
½
: X
Φ
n þ
1
2μ 1ÀΦ
ð
Þ
∂Θ
∂Δγ
K 4 = b
0
nþ1 Δγ þ b nþ1
ELSE
Elastic step ð Þ nþ1 = ð Þ
tr
nþ1 (Exit)
END IF
EXIT
214
5 Unified Mechanics of Thermo-mechanical Analysis