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Hysteresis loop and energy dissipation of

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Hysteresis loop and energy dissipation of

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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,P.R.China

    landc@mail.sdu.edu.cn xumingyu@163169.net

    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.

     1 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 P.R.China (Grant No.20030422046).

    1

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2 T h e b a s i c t h e o r i e s o f v i s c o el a s t i c s ol i d m o d e l s

The dissipation of mechanical energy, for one dimensional situation, can be written as t d,;!;t (1) W;!;;;;:,dt .

0 0

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;: (t ) is

t (5) (t );!;? K (t;;;, )( )d,; 0 or t (6) (t );! K (0) (t );?;? K ( ) (t;;;, )d . 0 Where K(t) is the relaxation modulus and overdots denote derivatives with respect to ,;[1,3].

K(t) is a convex function[1] or K(t) is a nonnegative monotone decreasing function and;; K (t ) is

monotone decreasing to zero. In addition,;, (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:

K (t ) G(t );!;.

K (0;? )

If the strain is step function, the developing process of stress is a function of time t and strain;, .

Using K ( , t ) to express this process, we have

K ( , t );! G(t )T (e) ( ) , G(0;? );! 1 , (22)

(e) ( ) is a function of strain and is called elastic response . where T

(t ) as a function of time t , with T being the period of;, (t ) .

-12Fig.1:

2

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3 H y s t e r e s i s l o o p a nd e n e r g y d i s s i pation under nonnegative inputted

strain

(t) , as shown in Fig.1, satisfies (5)~(8): In this section, the inputted strain

(t;? T );!;, (t ), (t;; [0,?;)), (5)

(6) (nT );! 0, (n;! 0,1,2,) ,

T (7) 2

~ 0, ; (nT , nT;? T ) t 2 ?; (8) t; (nT;? T , (n;? 1)T ). ? 0, 2 ?;

(t ) satisfies (2) or (3).

-12Moreover,

Using (3) we have t;?T d[ (t );;;: (t;? T )] d (t;? T;;;, ) . !;d(9) [;K ( )] dt dt t , b ; a0 2;!;;;? a0 ; in which a0; (0, T ) , 1;! t;? b1 and 2;! t;? b2 , then we T Let b1;! T 2 2 2

have

d (t;? T;;;, 1 )

dt

(T;; b1;?;)t );;;, (T;; b1 ) lim (10) )t;;0 )t d (T;? a0 ) 2 !; 0. dt

Similarly, we have d (T; a0 ) d (t;? T;;;, 2 ) 2 !; 0. (11) dt dt

Using (7) we obtain

d (T;? a0 ) d (T; a0 ) 2 2 ?; 0. dt dt

Therefore

3

     http://www.paper.edu.cn d (t;? T;;;, 1 ) d (t;? T;;;, 2 ) ?; 0. (12) dt dt

    Further because [;K (t )]( 0) is a monotone decreasing function,

    d[ (t );;;: (t;? T )]

    dt

     T t;? 2 d (t;? T;;;, ) d,;!;(13) [;K ( )] dt t t;?T d (t;? T;;;, ) d,;, 0. ?; [;K ( )] dt T t;?2

    What is more,

     (nT;? a0 );;;: (nT;? T;; a0 );, 0. (14) (see the proof of (32)). Using (13) and (14) we obtain that

    [ (nT;? a0 );;;: (nT;? T;? a0 )];; [ (nT;? T;; a0 );;;: (nT;? T;? T;; a0 )] (15) [ (nT;? a0 );;;: (nT;? T;; a0 )];; [ (nT;? T;? a0 );;;: (nT;? T;? T;; a0 )];, 0.

    In the strain(;, )-stress(;: ) coordinate system, equation (14) stands for the difference between loading

    and unloading curves (hysteresis loops width[4] in the (n+1)th period and at the point;, (t );!;, (a0 ) . 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 lim it ) of hysteresis loops ,when t;;;?; , is

     ( n1)T ( n1)T

     S lim it;! lim ;:d,;! lim ;:,dt n;;;n;;nT nT

    T (16)

    2 nT

     lim;?;? [;K ( )][ (nT;? T;; a0;;;, );;;, (nT;? a0;;;, )]dda0 . n;;;;;;;0 0

    In course of proving (16), we used the equation (24).

    S i t u a t i o n o f t h e s t r a i n b e i ng c o n s t a n t o n s o m e d o m a i n s 4

     Firstly, we prove that the Boltzmann superposition principle holds for the situation of the strain

    being constant on some domains.

    Proof. Letting T be the period of strain;, (t),T;! T1;? T2 ,;, (t);( 0(t;; (0,T1 )) and;, (t );! 0 ;;;;;;;;;;;;;;;;;;;;

    (t;; (T1,T )) . When t;; (0,T1 ) , the relation between;: (t ) and;, (t) is obviously that

     4

http://www.paper.edu.cn t )t (17) (t );! lim;; K (t;; j)t ){ [( j;? 1))t ];;;, ( j)t )}, )t;0 j;!0

(t ) satisfies (2) or (3). Namely, at present,

(t ) experiences a relaxation When t;; (T1 ,T ) , because;, (t) keeps constant on this domain,

process. This process can be written as

t )t

(t );! lim;; K (t;; j)t ){ [( j;? 1))t ];;;, ( j)t)} )t;0 j;!0 T1 (18) ;? K (t;;;, )( )d,; 0 t

;? K (t;;;, )( )d . 0

On the analogy of (18), we have ~()K (t; )( )d , t;; (nT , nT;? T1 ) ?; t;; (nT;? T1 , nT;? T ) ?()K (t; )( )d , (19) ?;

t t

;? K (t;;;, )( )d,;! K (0) (t );?;? K ( ) (t;;;, )d . 0 0

Similarly, (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: (t ) as a function of time t , with T being the period of;, (t ) , and t1;! T1 and t 2;; t1;! T2 .

This kind of strain, as shown in Fig.2, satisfies (5)~(7) and (20):

5

     http://www.paper.edu.cn ~ 0, t;; (0, t1 ) ??; (20) t;; (t1 , t 2 )

    ?, 0, t;; (t 2 , T ), ?

    ?where T;! T1;? T2 ( T1 and T2 are positive constants), t1;! T1 2 and t 2;; t1;! T2 . The second term

    of equation (20) stands for;, (t) being constant on [T1 2 ,T;; T1 2] , whose length is T2 .

    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.;, (t ) as a function of time t, with T being the period of;, (t ) and t1 T1 .

    This kind of strain, shown in Fig.3, satisfies (5),(6) and (21)~(23):

    ~, 0, t;; (0, t1 2) ?(21) (t )?;t;; (t1 2 , t1 ), ?? 0, (t );! 0, (22) t;; [t1 ,T ],

     (23) (a);!;, (t1;; a), a;; [0, t1 2],

where T;! T1;? T2 ( T1 and T2 are positive constants) and t1;! T1 .

     T ?;? then we have Letting ?;; [0, T 2], 1, 2;!;2

     (nT;? a);;;: (nT;? T1;; a)

     nT;? a ; a;;;, );;;, (nT;? a;;;, )]d,;!;;(24) [;K ( )][ (nT;? T1 0

     nT;?T1;;a

    ; a)d . ?;; [;K ( )] (nT;? T1 nT;? a

    Obviously, the second term of equation (24) is greater than zero. Further because

     6

     http://www.paper.edu.cn T T (25) 2 2

     T T (26) 2 2

     and using (5) and (23) we have

     T T 2 2 and T T 2 2

     Therefore

     (nT;? T1;; a;;;, 1 );;;, (nT;? a;;;, 1 );?;, (nT;? T1;; a;;;, 2 );;;, (nT;? a;;;, 2 );! 0. (27) Moreover, for convenience, letting

     f ( );!;, (nT;? T1;; a;;;, );;;, (nT;? a;;;, ), (28)

     T then, obviously, f (nT );! 0 . To study the sign of f ( ;;? ) , we divided this problem into two cases: 2

     T1;; 2a .

    Case1: T2 T1;? T2 T1; 2a a;;;,;!;;;When , we have 2 2

     f (nT;);! 0, nT;; (0, t ]. T T (29) 2 2 Further because f (a);, 0 and notice that the solutions of f ( );! 0 are nT , nT;; T 2, nT;;;

    ;? ,?; [0, ] , we arrive at T T (0, t], when;,;!;, 2;!;2 2

     T ;;? );, 0 . (30) f ( 2 );! f ( 2

     T1;; 2a . Case2: T2

    In this case, f ( );! 0 on ,;; (nT;? T1;; a;; T , nT;? a;; T1 ) .

     1 2;, 1 2 T T;? T T;? T;; 2a Because T1;; a , 2 2 2

    7

     http://www.paper.edu.cn T;? T T 1 and T;; (T1;; a);;; T2;? a; 1 2;!(T2;; T1;; 2a);, 0 , 2 2 2

     T T we have;;;; (T1;; a;; T , a;; T1 ) , namely ; (T1;; a, T2;? a) . 2 2

     T T ;?;;, a; T1;? T;! a;? T2 . Then we have 0;;;, 2;!;2 2

    Further because f (a);, 0, f (0);! 0 ,

     T ;;? );, 0 . we arrive at f ( 2

    In sum,

     T T T ;;? )];;;, [nT;? a;; ( ;;? )];, 0 . (31) f ( ;;? );!;, [nT;? T1;; a;; ( 2 2 2

    Moreover, using (27) and the monotonicity of [;K ( )]( 0) , the first term of (24) is also greater than

    zero. Therefore

     (nT;? a);;;: (nT;? T1;; a);, 0 . (32) The sign of equality holds, provided and only provided a;! T1 2 . Obviously, (14) holds when T2;! 0 (namely T;! T1 ) and a;! a0;; [0, T1 2) in equation (32).

     (t );;;: (t;? T );, 0 , we have Moreover, for

     t;?T1 d[ (t );;;: (t;? T )] d (t;?1;;;, T ) !; . d;(33) [;K (,;? T2 )] dt dt t

    On the analogy of (9), we can obtain

     d[ (t );;;: (t;? T )] 0 . (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;?T1 ;;, )d,;, 0 . ;;;;;;(35) (nT;? T1 );;;: (nT;? T );!; [K ( );; K (,;? T2 )] (nT;? T1 0

    Equation (35) implies that;: (t ) experience a relaxation process on t;; (nT;? T1 , nT;? T ) .

    Because;, (t );! 0 on this domain,;: (nT;? T );;;: (nT;? T1 ) , in the strain(;, )-stress(;: ) coordinate system, stands for the length of the overlap between the (n+1)th hysteresis loop and;: - coordinate axis.

    8

     http://www.paper.edu.cn )~(38).

     This length has close relation to;: (nT );;;: (nT;? T1 )

    on

    (nT

    T1

    )

    ,which

    is

    shown

    in

    (36

, the (n+1)th hysteresis loop's width

(nT );;;: (nT;? T1 );?;: (nT;? T1 );;;: (nT;? T );!;: (nT );;;: (nT;? T ) , (36)

where ( n1)T (nT );;;: (nT;? T );!;(37) [;K ( )] (nT;? T;;;, )d,;, 0 , nT and ( n1)T lim (38) [;K ( )] (nT;? T;;;, )d,;! 0 n;;; nT [4].

Namely, 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.

H y s t e r e s i s l o o p a n d e n e r g y d i s s i p a t i on u n d e r t h e l i n e a r 5

superposition of strains

For linear superposition of strains, it is easy to prove that the Boltzmann superposition principle

also holds.For 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.;, (t ) as a function of time t, with T being the period.

of;, (t ) ,;,1 (t ) and;, 2 (t ) and;, (t );!;,1 (t );?;, 2 (t ) . t2;; t1;! T1 , t3;; t 2;! T2 , t1;! T3 2 .

(t );!;,1(t );?;, 2 (t) Let satisfy (5),(6) and (39)~(41):

t (a);!;, (T;; a), (39) 2

9

a; [0, 1 ],

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