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Fractional Calculus in Mechanics of Solids	177

Using the Vi´ete theorem, from (41) and (46) we have

t−1 γ t−2 γ = τ1−γ τ2−γ (1 + n1 + n2),

τ1−γ τ2−γ = t−1 γ t−2 γ (1 − m1 − m2),

	t1−γ t2−γ		= 1 + n1	+ n2,	(47)
	−γ	−γ
	τ1	τ2
τ1−γ τ2−γ			= 1 − m1	− m2.	(48)
t−γ t−γ
1		2

Formulas (47) and (48) are generalized easily over the case of the operators deﬁned by relationships (23) and (24). In this case, they look like as follows:

	n		γ
	n					i=1
Q tj						i=1
i=1		τi				n
				= 1 +		X	(49)
	n
Q
j=1
	n		γ
j=1 tj						n
Q τi					− j=1
				= 1		X	(50)
	n
Q
i=1

4.Calculation of the Main Viscoelastic Operators

4.1. The Case of Constant Operator of the Bulk Extension-Compression e

For the majority of viscoelastic materials, the bulk modulus remains a constant value during the process of mechanical deformation, i.e.,

1 −e2ν	= 1 − 2ν∞ ,	(51)
E	E∞

where ν∞ is the nonrelaxed Poisson’s ratio, and νe is Poisson’s operator. From (51) with due account for (19) we have

νe = ν∞ +	1	(1 − 2ν∞) νε 3γ (τεγ ) .	(52)
	2

Further we will require the operators (1 + νe)−1 and (1 − νe)−1, which are determined considering (52) according to the above developed procedure, i.e.,

1 + ν∞ + 21 (1	− 2ν∞) νε 3γ	(τε )		1 + ν∞		− 3
	1		=	1	1	B	γ (t1γ ) ,	(53)
		γ	=		1	B	γ (t1γ ) ,	(53)

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178	Yury A. Rossikhin and Marina V. Shitikova

	1 − ν∞ − 21 (1 − 2ν∞) νε 3γ	(τε )		1 − ν∞		3
	1		=	1	1 + D		γ (t2γ ) ,	(54)
		γ	=		1 + D		γ (t2γ ) ,	(54)

where B, tγ1 , and D, tγ2 are yet unknown constants.

Multiplying the operators in the right-hand sides of Eqs. (53) and (54), respectively, by the denominator of the fraction in their left-hand side parts and considering (21), we are led to the following equations:

1 + ν∞

−

τε

− t1

3γ ε

−

− 2

1 + ν∞

τε − t1

3γ

(1 − 2ν∞) νε

τεγ

(τ γ )

B 1

(1 − 2ν∞) νε

t1γ

(t1γ )

= 0,

(55)

1 (1 − 2ν∞) νε

τεγ

–

1 (1 − 2ν∞) νε

t2γ

–

) = 0.

−

(τ

(56)

2 1 − ν∞

τεγ − tγ

1 − ν∞

τεγ

− tγ

1 + D

) + D 1 +

Vanishing to zero the terms in square brackets in (55) and (56), we could determine the unknown constants

B =

(1 − 2ν∞)νε

τεγ − t1γ

(57)

2(1 + ν∞) + νε(1 − 2ν∞)

τεγ

t−γ

= τ −γ

1 +

(1 − 2ν∞) νε

tγ

τεγ

(58)

2(1 + ν∞)

D =

(1 − 2ν∞)νε

τεγ

− t2γ

(59)

2ν

)

−

2(1

−

)

−

ν (1

−

τ γ

∞

−

2(1 − ν∞)

t−γ

= τ −γ

(1 − 2ν∞) νε

tγ =

τεγ

(60)

where

A =

2(1 + ν∞) + νε(1 − 2ν∞)

> 1,

C =

2(1 − ν∞) − νε(1 − 2ν∞)

< 1.

2(1 + ν∞)

2(1 − ν∞)

Now we could calculate operators

and

λ. Really, considering (19) and (53), (57),

(58), we ﬁnd

2(1 e

1 + ν∞

−

µ =

E∞

νε

(τε )

γ (t1 ) .

(61)

+ ν) 2

operators entering in (61) and considering (21), we have

Multiplying

t1γ 3γ

ε −

t1γ

3γ 1

∞ − ε − τεγ

τεγ

ε τεγ

tγ

µ = µ 1

ν 1

(τ γ )

B 1 + ν

(t ) , (62)

−

where µ∞ is the nonrelaxed shear modulus.

But according (57)

1 − Bτεγ (τεγ − tγ1 )−1 = 0,

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Fractional Calculus in Mechanics of Solids	179

and	1 + νε τεγ − t1γ		=	2(1 + ν∞)A,
B
		t1γ		3νε

and ﬁnally relationship (62) with due account for (63) takes the form

µ = µ∞	1 − 2(1 + ν∞)A		3γ	A			.
		3ν			τ	γ
e		ε				ε
e

To deﬁne operator e1, let us use the following formula:

e	(1 − 2 e e				1 −e2ν	− 3	e
λ =	E ν		=	1	E	1	E	.
	ν)(1 + ν)			3	e		1 + ν
	e	e			e		e

(63)

(64)

(65)

The ﬁrst operator in the right-hand side of (65), according to (51), is a constant, while the second one within an accuracy of a constant coincides with formula (64). Thus,

λ = 3 1

2ν∞ −

3 1 + ν∞ 1 − 2(1 + ν∞)A 3γ

E∞

1 E∞

3νε

τεγ

−

2(1 + ν∞)A

3γ

∞

λ = λ

1 +

(1 − 2ν∞)νε

τεγ

(66)

where λ∞ is the

nonrelaxed magnitude of the second Lame parameter.

In conclusion of this Subsection, we will calculate the operator

−e

ν2

= 2

1 + ν +

−

(67)

which within an accuracy of a constant coincides with the operator of cylindrical rigidity and is of frequent use in the problems dealing with mechanical behavior of thin bodies.

Substituting (19), (53), and (54) in (67), multiplying the operators involving in the enumerated relationships and using (21), we obtain

1 −eν2

1 − ν∞2

− 3

E∞

where

γ (t1 ) m2

γ (t2 ) ,

(68)

B(1 − ν∞)

D(1 + ν∞)

m1 =

m2 =

(69)

2 (1 − 2ν∞)

(1 − 2ν∞)

In order to ﬁnd an operator reverse to the operator (68), let us represent it in the form

1 − ν2

1 − ν∞2

+ n

(τ γ ) + n

(τ γ ) ,

(70)

3γ

E∞

where τ1γ , τ2γ , and n1, n2 are yet unknown constants which are determined from Eqs. (44) and (45). It is convenient to rewrite these equations in another form. Thus, the equation for deﬁning the values τiγ has the form

		x			x
1 + m1			+ m2			= 0, x = τiγ	(i = 1, 2),	(71)
1 + m1	γ	− x	+ m2	γ		= 0, x = τiγ	(i = 1, 2),	(71)
	t1	− x		t2	− x

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180	Yury A. Rossikhin and Marina V. Shitikova

or with due account for (69)

1 +

B(1 − ν∞)

1 D(1 + ν∞) x

= 0.

2 (1 − 2ν∞) (t2γ − x)

2 (1 − 2ν∞) (t1γ − x)

A set of equations for ﬁnding n1 and n2 is the following:

tγ

n1 +

tγ

n2 = 1,

−t

t2γ

τ1γ −t2γ

n1 +

τ2γ −t2γ

n2 = 1.

(72)

(73)

It could be shown that if substitute τεγ instead of x in (71) or (72) , then the following identity could be obtained:

τεγ τεγ

1 ≡ −m1 tγ1 − τεγ − m2 tγ2 − τεγ .

In order to ﬁnd the second root of Eq. (71) or (72), let us rewrite it in the form

(1 − m1 − m2) x2 − (tγ1 + tγ2 − m1tγ2 − m2tγ1 ) x + tγ1 tγ2 = 0.

Due to the Vi´ete theorem, if the ﬁrst root is x1 = τ1γ = τεγ , then the second root x2 satisﬁes the relationship

τεγ x2 =	t1γ t2γ	.

Calculations show that	1 − m1 − m2

t1γ t2γ	τε2γ

1 − m1 − m2 = 1 − νε .

(74)

(75)

= τ1γ

(76)

(77)

But τεγ (1 − νε)−1 = τσγ , and from (77) it follows that x2 = τ2γ = τσγ . Substituting the found τ1γ = τεγ and τ2γ = τσγ in the set of Eqs. (73), we obtain

2(1−ν∞)

−

2(1−ν∞)

= 1,

(1−2ν∞)ν

νσ

2(1+ν∞)ε

2 1+ν∞

−

(78)

n1 +

νσ

= 1,

(1−2ν∞)νε

whence it follows

(1 − 2ν∞)2νε

3νσ

n1 =

n2 =

(79)

4(1 − ν∞2 )

4(1 − ν∞2

)

Considering aforesaid, we have

4(1 − ν∞2

E∞

4(1 − ν∞2 )

)

1 − ν2

1 − ν∞2

(1 − 2ν∞)2νε

(τ

γ ) +

3νσ

(τ γ ) . (80)

Thus, we obtain two important mutually reciprocal operators (68) and (80) which are widely used in different applications.

Besides, two useful formulas have been found

t1γ t2γ = τεγ τσγ (1 − m1 − m2),	(81)
t1γ (1 − m2) + t2γ (1 − m1) = (τεγ + τσγ ) (1 − m1 − m2).	(82)

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Fractional Calculus in Mechanics of Solids	181

4.2.The Main Case

In the previous Subsection we have not take the second, or bulk, viscosity into account, in spite the fact that there are numerous evidence of its existence in scientiﬁc literature [36, 39, 74, 75].

One of the ﬁrst papers in the ﬁeld was published by Meshkov and Pachevskaya [39], wherein the attempt for considering the inﬂuence of the bulk relaxation on the internal friction phenomenon was carried out on the example of longitudinal harmonic vibrations of a three-dimensional hereditary elastic rod under the conditions of homogeneous deformation. A fractional exponential function has been used as a hereditary kernel. Investigation of the frequency dependence of the tangent of the phase shift between the stress and deformation, i.e., mechanical loss tangent, has revealed two peaks, namely, shear and bulk, in so doing the peak due to the shear deformation is ﬁve times larger than that resulting from the bulk deformation.

∞

ε1

Assume that operators µ and K are given in the form

−

= µ

(τ γ ) ,

(83)

∞

−

νε2

ε2

(τ

K = K

)

(84)

where νε1 = 1 − µ0µ−∞1 = 1 − τεγ1τσ−1γ , νε2 = 1 − K0K∞−1 = 1 − τεγ2τσ−2γ , µ0 and µ∞, K0 and K∞ are relaxed and nonrelaxed shear and bulk moduli, respectively, τε1 and τσ1, τε2

and τσ2 are relaxation and retardation times, the subscript l relates to shear relaxation and the subscript 2 to bulk relaxation.

Starting from formulas (83) and (84), reciprocal operators have the form

µ1e = µ−∞1 1 + νσ1 3γ (τσγ1) ,

1e = K∞−1 1 + νσ2 3γ (τσγ2) , K

where νσ1 = µ∞µ−0 1 − 1 = τσγ1τε−1γ − 1, and νσ2 = K∞K0−1 − 1 = τσγ2τε−2γ − 1. First we calculate the operator of compliance

J =

E =

3Kµ

= 3 µ + 3K .

1 1

1 3K + µ

in (87) yields

Substituting (85) and (86) e

e e

3γ

(τσγ2) ,

J = J∞

1 + 3 µ∞

3γ (τσγ1) + 9 K∞

1 νσ1 E∞

1 νσ2 E∞

3γ (τσγ1) + 3 (1 − 2ν∞) νσ2 3γ (τσγ2) .

J = J∞ 1 + 3 (1 + ν∞) νσ1

(85)

(86)

(87)

(88)

Let us consider longitudinal harmonic vibrations of a three-dimensional hereditary elastic rod under the conditions of homogeneous deformation when harmonic force acts inﬁnitely long. If rod’s axis coincides with the z-axis, then its longitudinal strain is

εzz = p0J eiωt,

(89)

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182	Yury A. Rossikhin and Marina V. Shitikova

where p0 and ω are the amplitude and frequency of the external force. Let us introduce the following operators:

3γ+ (τiγ ) x(t) = Z−∞						3γ (−s/τi) x(t − s)ds,
				t
	+		Z−∞			(γ)
Iγ x(t) =			t		(t − s)γ−1				x(s)ds.
Iγ x(t) =					(t − s)γ−1				x(s)ds.
Then according to (13)
				∞
	γ	iωt	X				n	−γ(n+1)		γ iωt
3γ+	(τi	) e	=		(−1)			τi		I+e ,
			n=0
but						∞
∞						∞
X						X
(−1)nτi−γ(n+1) I+γ eiωt =						(−1)n(iωτi)−γ(n+1) eiωt,
n=0						n=0

(90)

(91)

(92)

and the sum in the right-hand side could be interpreted as an inﬁnite decreasing geometrical progression with the denominator d = −(iωt)−γ . In other words,

∞

− −

eiωt

eiωt (−1)n(iωτi)−γ(n+1) =

(iωτi)−γ eiωt

[

(iωτi)−γ ]

1 + (iωτi)γ

n=0

and

3γ+ (τiγ ) eiωt =

eiωt

1 + (iωτi)γ

æi−γ + æiγ

+ 2 cos ψ !

æi−γ + æiγ

+ 2 cos ψ −

æi−γ

+ cos ψ

sin ψ

iωt

e ,

(93)

where æi

= ωτi, and ψ =

πγ.

Considering (93), the complex compliance J(iω) has the form

J(iω) = J

(ω) − iJ

(ω) =

3 J∞ (2(1 + ν∞)

æσ−1γ + æσγ1

+ 2 cos ψ

æσγ

1 + æε−1γ

+ [1 + (æσ1/æε1)γ ] cos ψ

−

æσ−2γ

+ æσγ2

+ 2 cos ψ

)

+ (1

2ν∞)

æσγ

2 + æε−2γ + [1 + (æσ2/æε2)γ ] cos ψ

−

(

−

)

æσ−1γ

+ æσγ1 + 2 cos ψ

æσ−2γ

æσγ2 + 2 cos ψ

J∞

2(1 + ν∞)

[(æσ1

/æε1)γ − 1] sin ψ

+ (1

2ν∞)

[(æσ2

/æε2)γ − 1] sin ψ

. (94)

Now knowing the value J(iω), it is possible to ﬁnd the tangent of the mechanical loss angle characterizing the internal friction in a viscoelastic rod

tan δ =	J00
tan δ =	J0 .	(95)

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Fractional Calculus in Mechanics of Solids	183

Meshkov and Pachevskaya [39] considered a numerical example with ν∞ = 0.3, µ0/µ∞ = K0/K∞ = 0.8, τε1/τε2 = 103, and ω = 1 in order to establish whether, in principle, a relaxation peak due to bulk deformation can occur. The ln æε2-dependence of tan δ and phase diagram for the compliance J00 = f (J0) were presented in [39] in Figures 1a and 1b, respectively, where the values of the fractional parameter γ were indicated by ﬁgures near the corresponding curves. The limiting value γ = 1 in Figure 1a of [39] indicates that the shear and bulk relaxations are described by the standard linear solid model, and it leads to clear separation of the two peaks. If the bulk and shear moduli have equal degrees of relaxation, they make unequal contributions to the total effect deﬁned by Poisson’s ratio ν∞, namely: the shear peak being about 5.5 times larger than the bulk one. Reduction in γ corresponds to broadening of the relaxation spectrum [41], and the difference between the peaks vanishes. The value γ = 0.5 may be considered in this case to correspond to the lower limit for the bulk effect, i. e., bulk relaxation is not seen for all γ < 0.5, although it exists. The possibility of observing bulk relaxation is thus dependent not only on the ratio of the relaxation times for shear and bulk stresses, but also on the parameter that characterizes the width of the relaxation spectrum.

Now we calculate the operator

ν =

3K − 2µ .

K + µ

We will proceed from the operator

−

3γ

ε2

13K + µ = (3K∞ + µ∞)

1 3γ (τε1) − m2

(τ γ ) ,

where m

−

, and m

+ ν

)ν

∞

ε1

∞

ε2

Operator inverse to operator (97) has the form

3K + µ

= 3K∞ + µ∞

1 + n1 3γ (tσγ1 ) + n2 3γ (tσγ2 ) ,

e γ

−1

and

where the values tσ1

tσ2

are deﬁned from the quadratic equation

(1 − m1 − m2)x2 − [τεγ1(1 − m2) + τεγ2(1 − m1)] x + τεγ1τεγ2 = 0,

(96)

(97)

(98)

(99)

while the values n1 and n2 are determined from the set of two equations with due account

for the magnitudes of tσγ

1 and tσγ

2 found from Eq. (99)

τεγ1

= −1,

τεγ1−tσγ 1

τεγ1−tσγ 2

(100)

τεγ2

= 1.

τε2

−tσ1

τε2

−tσ2

Then from (96) we ﬁnd

−

2 ) , (101)

ν = ν∞ 1 − s1 3γ (τεγ1) + s2 3γ (τεγ2) 1 + n1 3γ (tσγ

1 ) + n2 3γ (tσγ

where

3K∞νε2

2µ∞νε1

s1 =

s2 =

3K∞ − 2µ∞

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184	Yury A. Rossikhin and Marina V. Shitikova

	Opening the brackets in (101) with due account for (21) yields
	ν = ν∞ 1 + n1 + τεγ2		− tσγ1	− τεγ1		− tσγ1	3γ	(tσ1)
	e	s1n1tσγ1			s2n1tσγ1			γ
		s1n2tγ			s2n2tγ		3γ	(tσ2)
	+ n2 +	τεγ2	− tσγ2	− τεγ1		− tσγ2
			σ2			σ2		γ
	− 1	τεγ2	− tσγ1		τεγ2	− tσγ2	3γ	ε2
	s +	s1n1τεγ2		+	s1n2τεγ2			(τ γ )

+ s2 +

s2n1τεγ1

s2n2τεγ1

3γ

(τε1) .

(102)

τεγ1 − tσγ

In order to ﬁnd operator E, it is necessary to rewrite operator J in the form

J = J

1 + n

(τ

γ ) + n

(τ γ

) ,

(103)

e∞

σ1

σ2

where n1 =

2ν∞)νσ2.

+ ν∞)νσ1 , and n1 =

−

Then operator E takes the form

(tεγ1)

(tεγ2) ,

E = E∞

−

(104)

and

γ are determined from the following equation:

where values t

ε1

εe

n1x

n2x

1, x = tε i

(i = 1, 2),

τσγ1 − x

τσγ2 − x

(1 + n1 + n2)x2 − [τσγ1(1 + n2) + τσγ2(1 + n1)] x + τσγ1τσγ2 = 0,

(105)

and values m1 and m2 are deﬁned from the set of equations

τ γ

m2 = 1,

σ1

−t

σ1

ε1

σ1

ε2

(106)

τσγ2

τσγ2−tεγ1

τσγ2−tεγ2

m2 = 1.

In conclusion of this Subsection we will show how to deﬁne the operator proportional

to the operator of cylindrical rigidity. With this purpose we will use the operators

3K + µ

2(3K + µ)

E =

9Kµ

(107)

1 + ν

eK e

−

3Ke+

4µ

+ 4µ

3K e e

Then considering (107) we have

= µ +

9Kµ

= µ + s−1.

(108)

−e

2(1 + ν)

2(1 − ν)

3K + 4µ

First we will calculate the operator

s =

3K + 4µ

4 1

9Kµ

µ +

3 K

3K∞ + 4µ∞

3K∞ν

3µ∞ν

σ1

(τ

) +

σ2

(τ γ

(109)

σ1

1 e

σ2) .

9K∞µ∞

3K∞ + 4µ∞ 3

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Fractional Calculus in Mechanics of Solids

185

In order to ﬁnd an operator reverse to operator (109), let us rewrite it in the form

where

s = s∞ 1 + n1 3γ (τσγ1) + n2 3γ (τσγ2) ,

(110)

+ 4µ

3K∞νσ1

3µ∞νσ2

s∞ =

3K∞e

∞

n1 =

, n2 =

9K∞µ∞

3K∞ + 4µ∞

Then

s−1 =

9Kµ

= s∞−1

1 − m1 3γ (tεγ1 ) − m2 3γ (tεγ2) ,

(111)

4µ

where values t

γ3K e e

1 and tε2

are determined from Eq. (105), and values m1 and m2 are deﬁned

εe

from the set of Eqs. (106).

Finally, to calculate operator (108), it is necessary to take formulas (83) and (111) into

account. As a result we obtain

s∞

1 − µ∞s∞ + 1 3γ (τε1)

ν2

−e

µ∞s∞ + 1

µ∞s∞νε1

−

µ∞s∞

+ 1

ε1

− µ∞s∞

+ 1 3

ε2

γ (tγ

)

γ (tγ ) .

(112)

The reverse operator has the form

1 − ν2

s∞

1 + s1

(T γ ) + s2

(T γ ) + s3

(T γ ) .

(113)

γ∞s∞ +

σ1

σ2

σ3

e Tσ1

Tσ2

, and

Tσ3

could be found from the following equation:

Values

η1x

η2x

η3x

= −1,

τεγ1 − x

τεγ2 − x

τεγ3 − x

1 − η1 − η2 − η3

x3 + (τεγ2 + τεγ3) η1 + (τεγ3 + τεγ1) η2 + (τεγ1 + τεγ2) η3

x + τ

η1) + τ

η2) + τ

−

ε1 −

−

ε3

−

τε2(1 η3) x

ε2

ε3

ε1

ε1 γ γ

−γ

−τε1τε2

τε3

= 0,

(114)

where

η1 =

µ∞s∞νε1

η2 =

η3 =

τε2

= tε1

µ∞s∞ + 1

µ∞s∞ +

µ∞s∞ + 1

and values s1, s2, and s3 are determined from following set of equations

s1τ γ

s2τ

s3τ

= 1,

ε1

−T

−

ε1

σ1

ε1

σ2

σ3

−

s1τεγ2

s2τεγ2

εs3τεγ2

τ γ

−T

γ −T

τ γ −T

= 1,

ε2

σ1

ε2

σ2

ε2

σ3

s1τε3

s2τε3

s3τε3

= −1.

Applying the Vi´ete

τε3

−Tσ1

+ τε3−Tσ2

τε3−Tσ3

theorem to Eq.

(114), we have

γ γ γ

τεγ1τεγ2τεγ3

Tσ1Tσ2Tσ3

1 − η1 − η2 − η3

what completely corresponds to relationship (50) at n = 3.

τεγ3 = tγε2 ,

(115)

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