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F2 = c1 ( µ ’ µd ) . (5.94)
Elimination of µ™d , F1 and F2 from this set of equations leads to

F + „R F = c2 µ + ( c1 + c2 ) „R µ ,
™ (5.95)
with „R = c· /c1 the characteristic relaxation time. The force response to a step
µ( t) = µ0 H( t) in the strain yields
F( t) = F( t)hom + F( t)part = ±e’t/„R + c2 µ0 , (5.96)
with ± an integration constant to be determined from the initial conditions. Deter-
mining the initial condition at t = 0 for this problem is not trivial. It is a jump
condition with a discontinuous force F and strain µ at t = 0. A way to derive this
jump condition is by using the de¬nition of the time derivative:
F( t + t) ’ F( t)

F = lim . (5.97)
t
t’0
Let us take two time points a distance t apart, one point on the time axis left of
t = 0, which we call t = 0’ and one point on the right side of t = 0 which we
call t = 0+ . In that case Eq. (5.95) can be written as
F( 0+ ) ’ F( 0’ ) µ( 0+ ) ’ µ( 0’ )
F( 0) +„R = c2 µ( 0) +( c1 + c2 ) „R , (5.98)
t t
or
F( 0) t + „R ( F( 0+ ) ’ F( 0’ ) )
= c2 µ( 0) t + ( c1 + c2 ) „R ( µ( 0+ ) ’ µ( 0’ ) ) . (5.99)
89 5.4 Harmonic excitation of visco-elastic materials

F
µ
(c1 + c2)µ0




c2µ0


„R
t t

Figure 5.15
Response of the 3-parameter model to a step in the strain.




Because F( 0’ ) = 0 and µ( 0’ ) = 0 and the terms with t ’ 0 it
t vanish when
is found that

F( 0+ ) =( c1 + c2 ) µ0 , (5.100)

so ± = c1 µ0 . The solution is shown in Fig. 5.15:

F( t) = µ0 ( c2 + c1 e’t/„R ) . (5.101)

With Eq. (5.101) the step response G( t) is known. Using the Boltzmann integral
this leads to the general solution of Eq. (5.95):
t
c2 + c1 e’(t’ξ )/„R
F( t) = µ( ξ ) dξ .
™ (5.102)
’∞

There are several ways to determine the creep function. We can solve Eq. (5.95)
for a step in the force. This can be done by determining a homogeneous and a
particular solution as was done for the relaxation problem. However, it can also be
done by means of Laplace transformation of the differential equation. This leads
to an algebraic equation that can be solved. The result can be transformed back
from the Laplace domain to the time domain.
Instead of again solving the differential equation we can use the relation that
exists between the Creep function and the Relaxation function, Eq. (5.34). A
Laplace transformation of G(t) leads to

c2 ( s + 1/„R ) + c1 s
c1
c2
ˆ + =
G( s) = . (5.103)
s + 1/„R s( s + 1/„R )
s

With Eq. (5.34) the Laplace transform of J is found:
s + 1/„R
1
ˆ
J ( s) = = . (5.104)
ˆ [ ( c1 + c2 ) s + c2 /„R ] s
s2 G( s)
Fibres: time-dependent behaviour
90

c1 /c2 1
J ( s) = ’ +
ˆ . (5.105)
[ ( c1 + c2 ) s + c2 /„R ] c2 s
Back transformation leads to
1 c1
e’t/„K ,
J( t) = 1’ (5.106)
c2 + c1
c2
with „K =( c1 + c2 ) „R /c2 the characteristic creep time. It is striking that the char-
acteristic creep time is different from the characteristic relaxation time. To be
complete, the integral equation for force controlled problems is given:

t .
1 c1
e’(t’ξ )/„K F ( ξ ) dξ .
µ( t) = 1’ (5.107)
c1 + c2
c2
’∞


At the current point it is opportune to mention some terminology from system
dynamics. A linear system can be de¬ned by a transfer function. For a harmonic
excitation the transfer function is found by a Fourier transform of the original
differential Eq. (5.95):

F — ( ω) = E— ( ω) µ— ( ω) , (5.108)

with
c2 + ( c2 + c1 ) iω„R
E— ( ω) = . (5.109)
1 + iω„R
This can be rewritten as
1 + iω„K
E— ( ω) = c2 . (5.110)
1 + iω„R
In system dynamics it is customary to plot a Bode diagram of these functions. For
this we need the absolute value of E— ( ω):
1+( ω„K )2

|E ( ω) | = c2 . (5.111)
1+( ω„R )2
The phase shift φ( ω) is

φ( ω) = arctan( ω„K ) ’ arctan( ω„R ) . (5.112)

In our case „K > „R because c1 and c2 are always positive. The result is given in
Fig. 5.16.
The Storage and the Loss Modulus are the real and imaginary part of the
Complex Modulus E— ( ω):
1 + ω2 „K „R
E1 ( ω) = c2 (5.113)
1 + ω 2 „R
2
91 5.4 Harmonic excitation of visco-elastic materials

15



|E —(ω)| 10



5
10“1 100 101 102 103
ω
0.8

0.6

φ (ω) 0.4
0.2

0
10“1 100 101 102 103
ω

Figure 5.16
The Complex Modulus and phase shift for a standard linear model with two springs (c1 = 10,
c2 = 5) and one dashpot („R = 0.1).

15



E1(ω) 10



5
10“1 100 101 102 103
ω
6

4
E2(ω)
2

0
10“1 100 101 102 103
ω

Figure 5.17
The Storage and Loss Modulus for the standard linear model.

ω„K ’ ω„R
E2 ( ω) = c2 . (5.114)
1 + ω 2 „R
2

The result is given in Fig. 5.17. E1 has a similar shape as |E— ( ω) |, because the
asymptotic values for ω ’ 0 and ω ’ ∞ are the same. However, the slope of
Fibres: time-dependent behaviour
92

E1 is larger. The Loss Modulus E2 has its maximum at the point where the phase
shift is highest. This can be explained. At very high frequencies the dashpot has
an in¬nite stiffness and the behaviour of the standard linear model is dominated
by the two springs. At very low frequencies the in¬‚uence of the dashpot is small
and the behaviour is dominated by c2 . In these areas the mechanical behaviour is
like that of an elastic material.



5.5 Appendix: Laplace and Fourier transforms

In the current section a summary of the most important issues with regard to
Laplace and Fourier transforms will be given. Both transformations can be applied
to differential equations and transform these equations into algebraic equations. In
general the Fourier transform is used for periodic functions, the Laplace transform
is used for one-sided functions, meaning that the functions are zero up to a cer-
tain time and ¬nite after that time. In terms of visco-elasticity this means that the
Fourier transform is used as a tool to describe harmonic excitation and the Laplace
transform is used to describe creep and relaxation.
ˆ
The Laplace transform x( s) of a time function x( t) is de¬ned as

x( t) e’st dt.
x( s) =
ˆ (5.115)
0

The most important properties of Laplace transforms are:
• Laplace transform is a linear operation.
• When x( t) is a continuous function, the Laplace transform of the time derivative x( t)

of x( t) is given by
ˆ
x( t) = s x( s) ’ x( 0) ,
™ ˆ (5.116)

with x( 0) the value of the original function x( t) at time t = 0.
• Convolution in the time domain is equivalent to a product in the Laplace domain. Using
ˆ ˆ
two time functions x( t) and y( t) with Laplace transforms x( s) and y( s), the following
convolution integral I( t) could be de¬ned:

x( „ ) y( t ’ „ ) d„ .
I( t) = (5.117)
„ =’∞

In that case the Laplace transform of this integral can be written as
ˆ
I ( s) = x( s) y( s) .
ˆ ˆ (5.118)

• If a function x( t) has a Laplace transform x( s), then the Laplace transform of the
ˆ
function tn x( t), with n = 1, 2, 3, . . . can be written as
93 5.5 Appendix: Laplace and Fourier transforms

ˆ
dn x( s)
ˆ

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