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Hysteresis loop and energy dissipation of
viscoelastic solid models
1

Li Yan Xu Mingyu
Institute of Applied Math, School of Mathematics and System Sciences,
Shandong University,Jinan 250100,
landc@ xumingyu@

Abstract
In this paper the process of changing and the tendency of hysteresis loop and energy
dissipation of viscoelastic solid models were studied. One of the conclusions is that
under certain conditions, the sign of (65) is a sufficient and necessary condition for
judging the sign of the difference between dissipated energy in the (n+1)th period and
nth period. It has been proved that the Boltzmann superposition principle also holds
for inputted strain being constant on some domains. At the same time it was shown
that for the fractional-order Kelvin model and under the condition of quasi-linear
theory the above conclusions also hold.
Keywords: Viscoelastic solid models, relaxation modulus, Boltzmann superposition
principle, hysteresis loop, energy dissipation, quasi-linear principle, fractional
derivative theory.
1 Introduction
Under cyclic loading and unloading, viscoelastic materials exhibit hysteresis(a phase lag), which
is leading to a dissipation of mechanical energy. The mechanical damping relates to the storage and
dissipation of energy. And the energy stored over one full cycle is zero since the material returns to its
starting configuration. Therefore, the dissipated energy for a full cycle is proportional to the area within
the hysteresis loop. In other words, the area within the hysteresis loop represents an energy per volume
dissipated in the material, per cycle[1,2].
In this paper the process of changing and the tendency of hysteresis loop and energy dissipation of
viscoelastic solid models were studied. One of the conclusions is that under certain conditions, the sign
of (65) is a sufficient and necessary condition for judging the sign of the difference between dissipated
energy in the (n+1)th period and nth period. It has been proved that the Boltzmann superposition
principle also holds for inputted strain being constant on some domains. At the same time it was shown
that for the fractional-order Kelvin model and under the condition of quasi-linear theory the above
conclusions also hold.


The work was supported by the National Natural Science Foundation of China (Grant No.10272067) and by the
Doctoral Foundation of the Education Ministry of (Grant No.2).

1

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2 The basic theories of viscoelastic solid models
The dissipation of mechanical energy, for one dimensional situation, can be written as
tt
&
dt
. (1)
W=
∫
σ
d
ε
=
∫
σε
00
It consists of stored energy and dissipated energy. The energy stored over one full cycle is zero since
the material returns to its starting configuration. Therefore, the area within the hysteresis loop
represents an energy per volume dissipated in the material, per cycle [3]. Using the Boltzmann
superposition principle, the stress
σ
(t)
is
t
&
(
τ
)d
τ
(5)
σ
(t)=
∫
K(t−
τ
)
ε
0
or
&
(
τ
)
ε
(t−
τ
)d
τ
. (6)
σ
(t)=K(0)
ε
(t)+
∫
K
0< br>t
Where
K(t)
is the relaxation modulus and overdots denote derivatives with respect to
τ
[1,3].
&
(t)
is
K(t)
is a convex function[1] or
K(t)
is a nonnegative monotone decreasing function and
−K
monotone decreasing to zero. In addition,
ε
(t)
could be derived almost everywhere on
t∈(0,+∞)
.
Considering the nonlinear effect, we quote the quasi-linear theory[3].Using the quasi-linear theory,
we quote the definition of generalized relaxation modulus as follows:
G
(
t
)
=
K
(
t
)
.
K
(0
+
)
If the strain is step function, the developing process of stress is a function of time
t
and strain
ε
.
Using
K(
ε
,t)
to express this process, we have
K(
ε
,t)=G(t)T
(e)(
ε
)
,
G(0
+
)=1
, (22)
where
T
(e)
(
ε
)
is a function of strain and is called elastic response .

Fig.1:
ε
(t)
as a function of time
t
, with T being the period of
ε
(t)
.

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3 Hysteresis loop and energy dissipation under nonnegative inputted
strain
In this section, the inputted strain
ε
(t)
, as shown in Fig.1, satisfies (5)~(8):

ε
(t+T)=
ε
(t),(t∈[0,+∞)),
(5)
ε
(nT)=0,(n=0,1,2,L)
, (6)
ε
(a)=
ε
(T−a),(a∈[0,])
, (7)
⎧
>0,
⎪
&
(t)
⎨
ε
⎪<0,
⎩
Moreover,
σ
(t)
satisfies (2) or (3).
Using (3) we have
T
2
t ∈(nT,nT+
T
)
2
t∈(nT+
T
,(n+1)T) .
2
(8)
d
[
σ
(
t
)−
σ
(
t
+
T
)]
=
dt
Let
b
1
=
T
have
t+T
∫
t
&
(
τ
)]
d
ε
(
t
+
T
−
τ
)
d
τ
. (9)
[−
K
dt
−a
0
,b
2
=T
+a
0
，in which
a
0
∈(0,
T
)
,
τ
1
=t+b
1
and
τ
2
=t+b
2
, then we
222
d< br>ε
(
t
+
T
−
τ
1
)
dt
ε
(
T
−
b
1
+Δ
t
)−
ε
(
T
−
b
1
)
(10)

=l im
Δt→0
Δ
t
d
ε
(
T
+
a
0
)
2
=<0.
dt
Similarly, we have
T
d
ε
(t+T−
τ
2
)
d< br>ε
(
2
−a
0
)
=>0.
(11)
dtdt
Using (7) we obtain
d
ε
(
T
+a
0
)d
ε
(
T
−a
0
)
22
+=0.

dtdt
Therefore

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d
ε
(t+T−
τ
1
)d
ε
(t+T−
τ
2
)
+ =0.
(12)
dtdt
&
(t)](>0)
is a monotone decreasing function, Further because
[−K
d
[
σ
(
t
)
−
σ
(
t+T
)]
dt
t
+
T
2
=
+
What is more,
∫
t
t
+
T
t
+
&
(
τ
)]
d
ε
(
t
+
T
−
τ
)
d
τ
[−
K
dt
(13)
&
(
τ
)]
d
ε
(
t
+
T
−
τ
)
d
τ
<0.[−
K
∫
T
dt
2
σ
(nT+a
0
)−
σ
(nT+T−a
0
)>0.
(14)
(see the proof of (32)). Using (13) and (14) we obtain that
[
σ
(nT+a
0
)−
σ
(nT+T+a
0
)]−[
σ
(nT+T−a
0
)−
σ
(nT+T+T−a
0
)]
=[
σ
(
nT
+
a
0
)−
σ
(
nT
+< br>T
−
a
0
)]−[
σ
(
nT
+T
+
a
0
)−
σ
(
nT
+
T
+
T
−
a
0
)]>0.
(15)
In the strain(
ε
)-stress(
σ
) coordinate system, equation (14) stands for the difference between loading
and unloading curves (hysteresis loop’s width[4] in the (n+1)th period and at the point
ε
(t)=
ε
(a
0
)
.
Moreover, equations (13) ~ (15) indicate that the hysteresis loop's area decreases with increase of
period and is decreasing to a positive value. Namely, the energy dissipation for a full cycle decreases
with increase of period and is decreasing to a positive constant. The limit area(
S
limit
) of hysteresis
loops ,when
t→+∞
, is
(
n
+
1)
T< br>(
n
+
1)
T
n
→∞
S
limit
=lim
n
→∞
nT
∫
σ
d
ε< br>=lim
∫
σε
&
dt
nT
&
(
τ
)][
ε
(nT+T−a−
τ
)−
ε
(nT+a−
τ
)]d
τ
da.=lim
∫∫
[−K
000n
→∞
00
T
2
nT
(16)
In course of proving (16), we used the equation (24).
4 Situation of the strain being constant on some domains
Firstly, we prove that the Boltzmann superposition principle holds for the situation of the strain
being constant on some domains.
&
(t)≠0(t∈(0,T
1
))
and
ε
&
(
t
)=0

=T
1
+T< br>2
,
ε
(t∈(T
1
,T))
. When
t∈(0,T
1
)
, the relation between
σ
(
t
)
and
ε
(
t
)
is obviously that
Proof. Letting
T
be the period of strain
ε
(t),T

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t
Δt→
0Δt
σ
(t)=lim
∑
K(t−jΔt){
ε
[(j +1)Δt]−
ε
(jΔt)}
, (17)
j=0
Namely, at present,
σ
(
t
)
satisfies (2) or (3).
When
t∈(T
1
,T)
, because
ε
(
t
)
keeps constant on this domain,
σ
(
t
)
experiences a relaxation
process. This process can be written as
t
Δ
t
→
0
σ
(t)=l im
∑
K(t−jΔt){
ε
[(j+1)Δt]−
ε
(j Δt)}
j
=
0
T
1
Δ
t

&
(
τ
)d
τ
=
∫
K(t−
τ< br>)
ε
0
t
(18)
&
(
τ
)d
τ
.=
∫
K(t−
τ
)
ε
0
On the analogy of (18), we have
⎧
()K(t−
τ
)
ε
&
(τ
)d
τ
,
⎪
σ
(t)
=
⎨
⎪
()K(t−
τ
)
ε
&
(
τ
)d
τ
,
⎩
t
t∈(nT,nT+T
1
)
t∈(nT +T
1
,nT+T)
t
(19) < br>&
(
τ
)
ε
(t−
τ
)d
τ
.
&
(
τ
)d
τ
=K(0)
ε
(t)+
∫
K=
∫
K(t−
τ
)
ε
00
S imilarly, (19) holds for non-periodic functions. Therefore, the Boltzmann superposition principle
holds for the situation of the strain being constant on some domains.
4.1 Situation of the constant domain distributed symmetrically in a full period

Fig.2:
ε
(
t
)
as a function of time
t
, with T being the period of
ε
(
t
)
, and
t
1
This kind of strain, as shown in Fig.2, satisfies (5)~(7) and (20):
=T
1
and
t
2
−t
1
=T
2
.

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⎧
>0,
⎪⎪
&
(
t
)
⎨
=0,
ε
⎪
< 0,
⎪
⎩
where
T
t
∈(0,
t
1
)
t
∈(
t
1
,
t
2
)
(20)
t
∈(
t
2
,
T
),
=T1
+T
2
(
T
1
and
T
2
are positive constants),
t
1
=T
1
2
and
t
2
−t
1
=T
2
. The second term
of equation (20) stands for
ε
(
t
)
being constant on
[T
1
2,T−T
1
2]
, whose length is
T
2
.
Using (5)~(7) and (20) we can obtain that (13) and (15) also hold. Therefore, the hysteresis loop's
area decreases with increase of period and is decreasing to a positive value. Namely, the energy
dissipation for a full cycle decreases with increase of period and is decreasing to a positive constant.
4.2 Situation of the strain to be equal to zero on some domains

Figure 3.
ε
(
t
)
as a function of time t, with T being the period of
ε
(
t
)
and
t
1
This kind of strain, shown in Fig.3, satisfies (5),(6) and (21)~(23):
=T
1
.
⎧
>0,
⎪< br>&
(
t
)
⎨
ε
<0,
⎪
⎩
t
∈(0,
t
1
2)
t
∈(
t
1
2,
t
1
),
(21)
ε
(t)=0,t∈[t
1
,T],
(22)
a∈[0,t
1
2],
(23)
ε
(a)=
ε
(t
1
−a),
where
T=T
1
+T
2
(
T
1
and
T
2
are positive constants) and
t
1
=T
1
.
T
±
ξ
then we have
2
Letting
ξ
∈[0,
T2],
τ
1,2
=
σ
(nT
+
a)
−
σ
(nT
+
T
1
−
a)
nT+a
=
+
&
(
τ
)][
ε
(nT+T
∫
[−K
0
1
−a−
τ
)−
ε
(nT+a−
τ
)]d
τ
(24)
−a )d
τ
.
nT+T
1
−a
nT+a
&
(< br>τ
)]
ε
(nT+T
∫
[−K
1
Obvio usly, the second term of equation (24) is greater than zero. Further because

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ε
(nT+T
1
−a−
τ
1
)−
ε
(nT+a−
τ
1)=
ε
[T+T
1
−(+a+
ξ
)]−
ε(+a−
ξ
),
(25)
T
2
T
2
ε
(nT+T
1
−a−
τ
2
)−
ε(nT+a−
τ
2
)=
ε
[T+T
1
−(+a −
ξ
)]−
ε
(+a+
ξ
),
(26)
and using (5) and (23) we have
T
2
T
2
ε
[T+T
1
−(+a+
ξ
)]−
ε< br>(+a+
ξ
)=0

and
T
2
T
2
ε
[T+T
1
−(+a−
ξ
)]−
ε
(+a−
ξ
)=0.

Therefore
T
2
T
2
ε
(nT+T
1
−a−
τ
1
)−ε
(nT+a−
τ
1
)+
ε
(nT+T
1−a−
τ
2
)−
ε
(nT+a−
τ
2
)=0.
(27)
Moreover, for convenience, letting

f(
τ
)=
ε
(nT +T
1
−a−
τ
)−
ε
(nT+a−
τ
) ,
(28)
T
−
ξ
)
, we divided this problem into two cases:
2
then, obviously,
f
(
nT
)=0
. To study the sign of
f(
Case1:
T
2
≤T
1
−2a
.
=
T
1
+T
2
T
1
−
2
a
−
,
we have
22
f(nT−
T
)= 0,
2
nT−
T
∈(0,t]
. (29)
2
When
a−
τ
Further because
f
(
a
)>0
and notice that the solutions of
f
(
τ
)=0
are
{
τ
nT,nT−T2,nT∈

(0,t]
}
, when
τ
=
τ
2
=
TT
−
ξ
,
ξ
∈
[0,]
, we arrive at
22
f< br>(
τ
2
)=
f
(
T
−
ξ
) ≥0
. (30)
2
Case2:
T
2
>T
1
−2a
.
∈(nT+T
1
−a−T,nT+a−T
1
)
. In this case,
f
(
τ
)=0
on
τ
Because
T
T
1
+T
2
T1
+T
2
−
2
a
=>=T
1
−a,
222

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and
T−(T
1
−a)−
T+T
2
1
T
=T
2
+a−
1
=(T
2
−T
1
−2a)>0
,
222
we have
−
TT
∈(T
1
−a−T,a−T
1
)
, namely
∈(T
1
−a,T
2
+a)
.
2 2
TT
−
ξ
≤1
+T=a+T
2
.
22
Then we have
0≤
τ
2
=
Further because
f
(
a
)>0,
f
(0)=0
,
we arrive at
f
(
In sum,
T
−
ξ
)≥0
.
2
TTT
−
ξ
)=
ε
[
nT
+
T
1
−
a
−(−
ξ
)]−
ε
[
nT
+
a
−(−< br>ξ
)]≥0
. (31)
222
&
(
τ
)](>0)
, the first term of (24) is also greater than Moreover, using (27) and the monotonicity of
[−K
f
(
zero. Therefore
σ
(nT+a)−
σ
(nT+T
1
−a)≥0
. (32)
The sign of equality holds, provided and only provided
a=T
1
2
. Obviously, (14) holds when
T
2
=0
(namely
T=T
1
) and
a=a
0
∈[0,T
1
2)
in equation (32).
Moreover, for
σ
(
t
)−
σ
(
t
+
T
)>0
, we have
t+T1
d
[
σ
(
t
)−
σ
(
t< br>+
T
)]
=
dt
On the analogy of (9), we can obtain
∫
t
&
(
τ
+< br>T
)]
d
ε
(
t
+
T
1
−
τ
)
d
τ
. (33)
[−
K
2
dt
d
[
σ
(
t
)
−
σ
(
t
+
T
)]
<0< br>. (34)
dt
Therefore, the hysteresis loop's width decreases with increase of period and is decreasing to a
positive value. Namely, the hysteresis loop's area decreases with increase of period and is decreasing to
a positive value. In other words, the energy dissipation for a full cycle decreases with increase of period
and is decreasing to a positive constant.
In addition, using (3) we have
nT+T
1
σ
(nT+T
1
)−
σ
(nT+T)=
&
(
τ
)−K
&
(
τ
+T)]
ε
(nT+T
∫
[K
2
0
1
−
τ
)d
τ
<0
. (35)
Equation (35) implies that
σ
(
t
)
experience a relaxation process on
t∈(nT+T
1
,nT+T)
.
Because
ε
(
t
)=0
on this dom ain,
σ
(nT+T)−
σ
(nT+T
1
)
, in the strain(
ε
)-stress(
σ
) coordinate
system, stands for the length of the overlap between the (n+1)th hysteresis loop and
σ
- coordinate axis.

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This length has close relation to
σ
(nT)−
σ
(nT
on
ε
(nT
+T
1
)
, the (n+1)th hysteresis loop's width
+T
1
)
,which is shown in (36)~(38).
σ
(nT)−
σ
(nT+ T
1
)+
σ
(nT+T
1
)−
σ
(nT+ T)=
σ
(nT)−
σ
(nT+T)
, (36)
where
(
n+
1)
T
σ
(nT) −
σ
(nT+T)=
and
(
n+
1)
T
nT
&
(
τ
)]
ε
(nT+T−
τ
)d
τ
>0
, (37)
∫
[−K
lim
n→∞
nT
&
(
τ
)]< br>ε
(nT+T−
τ
)d
τ
=0
(38)
∫
[−K
[4].
Namely,
σ
(nT+T)−
σ
(nT+T
1
)
is monotone increasing to
lim[
σ
(nT)−
σ
(nT+T
1
)]
with
n→∞
the increase of
n
. Therefore, hysteresis loops tend to a closed loop. The area within this closed loop is
proportional to the minimum dissipated energy for a full cycle.
5 Hysteresis loop and energy dissipation under the linear
superposition of strains
For linear superposition of strains, it is easy to prove that the Boltzmann superposition principle
also convenience, we use the superposition of two strains to discussthis problem, in this
section.
5.1 Situation of the derivative of strain being zero on some domains

Figure 4.
ε
(
t
)
as a function of time t, with T being the period.
of
ε
(t)
,
ε
1
(t)
and
ε
2
(t)
and
ε
(t)=
ε
1
(t)+
ε
2(t)
.
t
2
−t
1
=T
1
,
t
3
−t
2
=T
2
,
t
1
=T
3
2
.
Let
ε
(t)=
ε
1
(t)+
ε
2
(t)
satisfy (5),(6) and (39)~(41):
ε
(
a
)
=
ε
(
T−a
),
t
a∈
[0,
1
],
(39)
2

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ε
(
t
1
+b
)
=
ε
(
t
2
−b
),
ε
(t)=
ε
(t
2
)=< br>ε
(t
3
),
where
b∈
[0,
t
2
−t
1
],
(40)
2
t∈[t
2
,t
3
],
(41)
⎧
⎪
⎪
ε
(
t
)
ε
1
(
t
)
=
⎨
⎪
ε
(
t
)
=
ε
(
t
)
=
ε
(
t
)
23
⎪
⎩
⎧
⎪
⎪
0
ε
2
(
t
)=
⎨
⎪
ε
(
t
)
−
ε
(
t
)
2
⎪
⎩
with
T
t
∈
(0,
t
1
)
∪
(
t
2,
T
)
t
∈
[
t
1
,
t2
],
(42)
t< br>∈
(0,
t
1
)
∪
(
t
2
,
T
)
t
∈
[
t
1
,
t
2
],
(43)
=T
1
+T
2
+T
3
(
T1
,T
2
and
T
3
are positive constants).
ε
(t),
ε
1
(t)
and
ε
2
(t)
are shown in
Fig.4. Obviously,
ε
1
(t)
and
ε
2
(t+t
1
)
satisfy (5)~(7) and (20) and (5)~(6) and (21)~(23)
respectively. Letting
&
(
τ
)
ε
(t−
τ
)d
τ
,
&
i
(
τ< br>)d
τ
=K(0)
ε
i
(t)+
∫
K
σ
i
(t)=
∫
K(t−
τ
)
ε
i00
(
n+
1)
T
tt
i=1,2,
(44)
W
ni
(t)=
Then we obtain that (
n+
1)
T
nT
∫
σ
i
&
i
(t)dt,(t)
ε
i=1,2
. (45)
W
n
(t)=
nT
∫
σ
(t)
ε
&
(t)dt
nT+T
3
nT
⎛
=W
n
1
+W
n
2
+
⎜
⎜
⎝
Using
∫
⎞
&
2
(t)dt+
∫
σ
2
( t)
ε
&
1
(t)dt.+
∫
⎟
σ
1(t)
ε
⎟
nT+T−T
3
⎠
nT+T
3nT+T
nT+T
1
(46)
lim[
σ
1
(nT+a)−
σ
1
(nT+T−a)]=l im[
σ
1
(nT+T+a)−
σ
1
(nT+T−a)]< br>
n→∞n→∞
and the same way as is used in arriving at (37), we have
lim[
σ
1
(n T+a)−
σ
1
(nT+T−a)]>0
. (47)
n→∞
Using (32) we obtain
lim[
σ
2
(nT+T
3
+b)−
σ
2
(nT+T
1
−b)]>0
. (48)
n→∞

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Moreover, in the same way as (13), we have
d
[
σ
(
t
)
−
σ
(
t
+
T
)]
<0
. (49)
dt
Therefore, the last two terms of (46)
nT+T
1
⎛
nT+T
3
nT+T
⎞< br>⎜
&
2
(t)dt+
∫
σ
2
(t)
ε
&
1
(t)dt>0
. (50)
+
∫
⎟
σ
1
(t)
ε
∫
⎜
nTnT+T−T
⎟
nT+T
33
⎠
⎝
Nam ely,
W
n
>W
n1
+W
n2
.
What is more, (49) and (50) imply that the hysteresis loop's area decreases with increase of period
and is decreasing to a positive value. Namely, the energy dissipation for a full cycle decreases with
increase of period and is decreasing to a positive constant.
5.2 Situation of strain's value being real

Figure 5.
ε
(
t
)
as a function of time t, with T being the period of
ε
(
t
)
and
t
1
Letting the inputted strain satisfy (5),(6) and (51)~(54):
=T
1
.
ε
(
a
)
=
ε
(
t
1
−a
),(
a∈
[0,
1
]),
(51)
ε
(
t
1
+b
)
=
ε
(
T−b
),(
b∈
[0,
⎧
>0,
⎪
ε
(
t
)
⎨
⎪
<0,
⎩
t
∈(0,
t
1
)
(53)
t
2
T
2
]),
(52)
2
t
∈(
t
1
,
T
),
⎧
>0,
⎪
⎪
ε
(
t
)
⎨
⎪< br><0,
⎪
⎩
tT
t
∈(0,
1
)∪(
T
−
2
,
T
)
22
(54)
t
1
T
2
t
∈(,
T
−),< br>22

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and
⎧
=
ε
(
t
),
⎪
ε
1
(
t
)
⎨
⎪
=0,
⎩
⎧
=0,
⎪
ε
1
(
t
)
⎨
⎪
=
ε
(
t
),
⎩
with
T
t
∈[0,
t
1
]
(55)
t
∈(
t
1
,
T
],
t
∈[0,
t
1
]
(56)
t
∈(
t
1
,
T
],
=T1
+T
2
(
T
1
and
T
2
are positive constants) and
t
1
=T
1
, as shown in Fig.5. Continuous
functions
ε
1
(t)
and
ε
2
(t)
are nonnegative and nonpositive parts of
ε
(
t
)
, respectively.
There exist two ways to find the natures of hysteresis loops and energy dissipation in this kind of
inputted strain. The one is to calculate directly the hysteresis loop's area and the other is to use the
superposition of
ε
1
(t)
and
ε
2
(t)
. With the second method, we can see clearly the interaction between
ε
1
(t)
and
ε
2
(t)
. What is more, using the two methods, we prove out the same conclusions about
hysteresis and energy dissipation. Therefore, we choose the second method to study the changing of
hysteresis loops.
Using (44) and the proofs of (32) and (48), we have

σ
1
(nT
Obviously,
σ
1
(nT< br>+T−b)−
σ
1
(nT+T
1
+b)>0.
(57)
+T−b)−
σ
1
(nT+T
1
+b)
is an increasing function of
n
. Moreover,
σ
2
(nT+a)−
σ
2
(nT+T
1
− a)>0.
(58)
Using (44) we have:
σ
(nT
+
T
+
a)
−
σ
(nT
+
T
+
T< br>1
−
a)
−
[
σ
(nT
+
a)−
σ
(nT
+
T
1
−
a)]
=
σ
1
(nT+T+a)−
σ
1
(nT+T+T
1
−a)−[
σ
1
(nT+a)−
σ
1
(nT+T
1
−a)]
(59)
+
σ
2
(nT+T+a )−
σ
2
(nT+T+T
1
−a)−[
σ
2
(nT+a)−
σ
(nT+T
1
−a)].
Then substituting (56) and (58) into (59) we have:
σ
2
(nT+T+a)−
σ
2
(nT+T+T
1
− a)−[
σ
2
(nT+a)−
σ
(nT+T
1
−a )]

nT+T
2
=
nT
&
(
τ
+a)−K
&
(
τ
+T
∫
[K
1
−a)]
ε
2
(nT+T−
τ
)d
τ
>0
. (60)
In terms of Boltzmann superposition principle we have

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σ
1
(nT+a)−
σ
1
(nT+T
1
−a)
nT+a
=
Therefore,
&
(
τ
)−K
&
(
τ
+T−2a)]
ε
(nT+a−< br>τ
)d
τ
−
11
∫
[K
0
T
1
−2a
&
(
τ
)
ε
(nT+T−a−
τ
)d
τ
.
11
∫
K
0
(61) < br>σ
1
(nT+T+a)−
σ
1
(nT+T+T
1−a)−[
σ
1
(nT+a)−
σ
1
(nT+T
1
−a)]

nT+T
=
nT+T
2
&
(
τ
+a)−K
&
(
τ
+T
∫
[K
1
−a)]
ε
2
(nT+T−
τ
)d
τ
<0
. (61)
Substituting (60) and (62) into (59) yields σ
(nT
+
T
+
a)
−
σ
(nT+
T
+
T
1
−
a)
−
[
σ< br>(nT
+
a)
−
σ
(nT
+
T
1< br>−a)]
(
n
+
1)
T
=
def
.
nT
&
(
τ
+a)−K
&
(
τ
+ T
∫
[K
1
−a)]
ε
(nT+T−
τ
) d
τ
(63)
=M(a,n).
Similarly, we have
σ
(nT
+
2T
−
b)
−
σ
(nT
+
T
+
T
1
+
b)
−
[
σ
(nT
+< br>T
−
b)
−
σ
(nT+T
1
+b)]
(
n+
1)
T
=
nT
&
(
τ
+ T−b)−K
&
(
τ
+T
∫
[K
1
+b) ]
ε
(nT+T−
τ
)d
τ
(64)
=N(b,n).
It is obvious that the sign of
M
(
a
,
n
)
or
N
(
b
,
n
)
depends only on
K
(
t
)
and
ε
(
t
)
. That is to say,
the dissipated energy for a full cycle is unnecessary monotone function of
n
. The tendency of change
of dissipated energy for full cycles is determined by (65)
T
1
2< br>T
2
2
∫
M(a,n)da+
∫
N(b,n)db< br>. (65)
00
Equation (65) stands for the difference between dissipated energies in the (n+1)th period and nth period.
Under the conditions of (5),(6) and (51)~(54), the sign of (65) is a sufficient and necessary condition
for judging the sign of the difference between dissipated energy, per volume, in the (n+1)th period and
nth period.
Moreover, because of the precondition,
T
1
2
T
2
2
∫
M(a,+∞)da+
∫
N(b,+∞)db=0
. (66)
00
holds for arbitrary strain[4].

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Obviously, (65) converts into (15) with
T
2
For example, letting strain
=0
.
obtained that (65) is greater than zero for arbitrary
n
∈(0,1,2,L)
. Namely, the dissipated energy for a
full cycle increases with increase of period.
On the real domain and letting
t
1
ε
(
t
)=sin(
t
)
, then we have
ε
1
(
t
)
=[sin(t)+sin(t)],

2
T
1
ε
2
(
t
)
=[sin( t)−sin(t)]
and
T
1
=T
2
=
. Substituting
ε
(t)，
ε
1
(t)
and
ε
2
(t)
into (65), it is
2
2
→+∞
, we have,
t→+∞
lim
σ
(t)=Asin(t)+Bcos(t)
, (67)
where
&
(
τ
)cos(
τ
)d
τ
]
, (68)
A=lim[K(0)+
∫
K
t
→+∞
0
t
&
(
τ
)]sin(
τ
)d
τ
}. (69)
B=lim{
∫
[−K
t
→+∞
0
t< br>Using the nature of
K
(
t
)
yields
&
(
τ
)]sin(
τ
)d
τ}>0
. (70)
B=lim{
∫
[−K
t
→+∞
0
t< br>After a straight calculation we have
(A
2
+B
2
)
ε
2
+
σ
2
−2A< br>εσ
=B
2
. (71)
Obviously, in
ε
−
σ
coordinate system, (71) is an inclined ellipse whose center is the origin(just like
discuss in [1]. The area
π
B
represents the most energy per volume dissipated in the material, full cycle.
Namely, the dissipated energy for full cycle is monotone increasing to a positive value, which is
proportional to
π
B
[5~7].
6 Applying of fractional Kelvin model and quasi-linear theory
Usage of the quasi- linear theory corresponds with substituting
T
and
K
(
t
)
respectively. Moreover, provided
T
(
e)
(
e
)
[
ε
(t)]
and
K
(
ε
,
t
)
into
ε
(
t
)

[
ε
]
satisfies (72) and (73):
⎧
>
0,
⎪
⎪
⎪
(e)
T
[
ε
]
⎨
=
0,< br>⎪
⎪
<
0,
⎪
⎩
ε
>
0
ε
=
0
ε
<
0,
(72)

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dT
(
e
)
[
ε
]
>0,
(73)
d
ε
all the above conclusions also hold.
Using the Riemann-Liouville derivative theory[8,9] and fractional Kelvin model, the relaxation
modulus[10] is
α
τ
ε
α
−
τ
σ
t
α
K
α
(t)=E
R< br>[1−E(−())]H(t)
, (74)
α
τ
ε
τ
ε
α
where
α
(0<
α
≤1)
is the fractal order of evolution,
E
α
(−(
t
τ
σ
))=
∑
α
n
=
0
∞
[−(
t
τ
σ
Γ(n
α
+1)
)
α
]
n

is Mittag-Leffler function and
H
(
t
)
is unit step function.
Because
K
α
(t)
is a complete monotone function[11],
K
α
(t)
is a convex function or
K
α
(t)
is a
&
(t)
is monotone decreasing to zero. Therefore, nonnegative monotone decreasing functions and
−
K
α
substituting
K
α
(t)
into
K
(
t
)
, the above conclusions also hold.
7 Results and discussion
In this paper, we discuss the process of changing and the tendency to one-dimensional viscoelastic
solid models' hysteresis loops and energy dissipation. We assume that
K
(
t
)
is a convex function or
&
(t)
is monotone decreasing to zero.
K
(
t
)
is a nonnegative monotone decreasing function and
−K
One of our conclusions is that under the conditions of (5),(6) and (51)~(54), (65)
≥0
implies that the
dissipated energy in the (n+1)th period is greater or equal to it in the nth period, as shown in the
example of section 5.2. On the other hand, (65)
<0
implies that the dissipated energy in the (n+1)th
period is less than it in the nth period, as discusses in sections 3~5.1. This is a sufficient and necessary
condition. Moreover, because of the precondition, (66) holds for arbitrary strain. We have proved that
the Boltzmann superposition principle also holds for inputted strain being constant on some domains.
At the same time, we prove that for the fractional-order Kelvin model and under the condition of
quasi-linear theory the above conclusions also hold.
8 acknowledgements
This work was supported by the National Natural Science Foundation of China (Grant
No.10272067) and by the Doctoral Foundation of the Education Ministry of (Grant
No.2).
References
[1]
Roderic S. Lakes(1999),Viscoelastic Solids,CRC Press, Boca Raton London New York Washington,
D.C.

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[2] G. Carbone and B.N.J. Persson(2005), Crack motion in viscoelastic solids: The role of the flash
temperature. Eur. Phys. J. E 17, 261{281}
[3]
hanics:Mechanical(1993), Properties of Living Tessues.2
nd
ed. New York:
Springer.

[4] Li Yan and Xu Mingyu(2006), Hysteresis and preconditon of viscoelastic solid models.

Mech Time- Depend
Mater (2006) 10:113–123.
[5] Anna Vainchtein and P. Rosakis(1999), Hysteresis and Stick-Slip Motion of Phase Boundaries in Dynamic
Models of Phase Transitions. J. Nonlinear Sci. Vol. 9: pp. 697–719.
[6] Anna Vainchtein(1999), Dynamics of Phase Transitions and Hysteresis in a Viscoelastic Ericksen’s Bar on an
Elastic foundation.

Journal of Elasticiy 57:243-280.
[7] Ian D. Aiken, Douglas K. Nims, Andrew S. Whittaker, James M. Kelly(1993), Testing of Passive Energy
Dissipation Systems. EARTHQUAKE SPECTRA, VOL. 9, NO. 3, EARTHQUAKE ENGINEERING
RESEARCH INSTITUTE CALIFORNIA AUGUST.
[8] I. Podlubny(1999), Fractional Differential Equations.

Academic Press, San Diego- Boston-New Yor –London
-Tokyo-Toronto.
[9]
and (1983),A theoretical basis for the application of fractional calculus to
viscoelasticity, Journal of Rheology 27,201-210.

[10]
Alan Freed , Kai Diethelm,et al.(2002), Fractional-Order Viscoelasticity(FOV):Constitutive
development using the fractional calculus:First annual -2002-211914.

[11]
Kenneth ,Stefan G. Samko(2001).Completely monotonic Spec.
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