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„:
corresponding to a viscous response, while for t
F
µ≈ , (5.73)
c
re¬‚ecting a purely elastic response.


5.4 Harmonic excitation of visco-elastic materials

5.4.1 The Storage and the Loss Modulus

In this section some methods will be described that can be used to calculate the
response of a visco-elastic material for different excitations. The section is aimed
at closed form solutions for the governing equations. Fourier and Laplace trans-
forms and complex function theory are used. First, the methods will be outlined
in a general context. After that, an example of a standard linear model will be
discussed.
In the previous section ¬rst-order differential equations appeared to describe
the behaviour of simple visco-elastic models, however, in general, a linear visco-
elastic model is characterized by either a higher-order differential equation (or a
set of ¬rst-order differential equations):
dN µ
dM F
dF dµ
p0 F + p1 + · · · + pM M = q0 µ + q1 + · · · + qN N , (5.74)
dt dt dt dt
or an integral equation:
t

µ( t) = J( t ’ ξ ) F( ξ ) dξ (5.75)
0

t
F( t) = G( t ’ ξ ) µ( ξ ) dξ .
™ (5.76)
0
Both types of formulation can be derived from each other.
When a model is used consisting of a number of springs and dashpots, the creep
and relaxation functions can be expressed by a series of exponential functions. It is
Fibres: time-dependent behaviour
84

said, that the functions form a discrete spectrum. It is possible, and for biological
materials sometimes necessary [8], to use continuous functions for G( t) and J( t)
thus establishing a more general identi¬cation than the differential formulation
with limited M and N.
To determine the response to some arbitrary force or strain history several solu-
tion methods are available. In the case of a differential model a usual way is to
determine the homogeneous solution and after that a particular solution. The gen-
eral solution is the summation of both. This is a method that is applied in the time
domain. Another approach is to use Laplace transforms. In this case the differen-
tial equation is replaced by an algebraic equation in the Laplace domain which
usually is easy to solve. This solution is then transformed back into the time
domain (often the harder part). This approach is usually used for functions that
are one-sided, meaning that the functions are zero up to a certain time and ¬nite
after that time.
The result of the integral formulation can, for the case of discrete spectra,
be considered as the general solution of the associated differential equation and
sometimes it is possible to derive closed form expressions for the integrals. This
depends strongly on the spectrum and the load history. If no closed form solutions
are available numerical methods are necessary to calculate the integrals, which is
usually the case for realistic loading histories.
Closed form solutions can only be generated for simple loading histories.
Strictly speaking the above methods are only applicable for transient signals (zero
for t < 0 and ¬nite for t ≥ 0). For examples see Fig. 5.13.
A frequently used way of excitation in practice is harmonic excitation. In that
case the applied loading history has the form of a sine or cosine. Let us assume
that the prescribed strain is harmonic according to

µ( t) = µ0 cos( ωt) . (5.77)

In case of a linear visco-elastic model the output, i.e. the force, will also be a
harmonic:

F( t) = F0 cos( ωt + δ) , (5.78)

or equivalently:

F( t) = F0 cos( δ) cos( ωt) ’ F0 sin( δ) sin( ωt) . (5.79)

This equation reveals that the force can be decomposed into two terms: the
¬rst with amplitude F0 cos( δ) in phase with the applied strain (called the elas-
tic response), the second with amplitude F0 sin( δ), which is 90—¦ out of phase with
the applied strain (called the viscous response). Eq. (5.79) can also be written as
85 5.4 Harmonic excitation of visco-elastic materials


f (t )

step



t
f (t )
block


t

f (t )
ramp



t

Figure 5.13
Some ˜simple™ functions that can be used as loading histories and for which closed form solutions
exist for linear visco-elastic models.



F( t) = µ0 E1 cos( ωt) ’ µ0 E2 sin( ωt) , (5.80)
with: E1 =( F0 /µ0 ) cos( δ) the Storage Modulus and E2 =( F0 /µ0 ) sin( δ) the
Loss Modulus. These names will become clear after considering the amount of
energy (per unit length of the sample considered) dissipated during one single
loading cycle. The necessary amount of work for such a cycle is
2π/ω
W= F µdt

0
2π/ω
=’ [µ0 E1 cos( ωt) ’ µ0 E2 sin( ωt) ] µ0 ω sin( ωt) dt
0
= πµ0 E2 .
2
(5.81)
It is clear that part of the work is dissipated as heat. This part, given by Eq. (5.81) is
determined by E2 , the Loss Modulus. During loading the E1 related part of F also
contributes to the stored work, however this energy is released during unloading.
That is why E1 is called the Storage Modulus.


5.4.2 The Complex Modulus

In literature on visco-elasticity the Complex Modulus is often used, which is
related to the Storage and Loss Modulus. To identify this relation a more formal
way to study harmonic excitation is pursued. In the case of a harmonic signal the
Boltzmann integral for the relaxation function can be written as
Fibres: time-dependent behaviour
86

t
F( t) = G( t ’ ξ ) µ ( ξ ) dξ .
™ (5.82)
’∞

The domain of the integral in Eq. (5.82) starts at ’∞ because it is assumed that
at the current time t the harmonic strain is applied for such a long time, that all
effects from switching on the signal have disappeared (also meaning that using a
Laplace transform for these type of signals is not recommended). The time t in the
upper boundary of the integral can be removed by substitution of ξ = t ’ s in Eq.
(5.82), thus replacing the integration variable ξ by s:

0
F( t) = ’ G( s) µ ( t ’ s) ds =
™ G( s) µ ( t ’ s) ds.
™ (5.83)
∞ 0
Substitution of (5.77) into this equation leads to

F( t) = µ0 cos( ωt) ω G( s) sin( ωs) ds
0

’ µ0 sin( ωt) ω G( s) cos( ωs) ds . (5.84)
0

The terms between brackets [ ] are only functions of the frequency and not of
the time. These terms are solely determined by the type of material that is being
considered and can be measured. We can write Eq. (5.84) as

F( t) = µ0 E1 ( ω) cos( ωt) ’ µ0 E2 ( ω) sin( ωt) , (5.85)

where the formal de¬nitions for the Storage and Loss Modulus:

E1 ( ω) = ω G( s) sin( ωs) ds (5.86)
0


E2 ( ω) = ω G( s) cos( ωs) ds, (5.87)
0
can be recognized.
In the case of harmonic excitation it is worthwhile to use complex function
theory. Instead of (5.77) we write

µ( t) = Re{µ0 eiωt }. (5.88)

The convolution integral for the force, Eq. (5.82) can be written as
∞ ∞
F( t) = G( t ’ ξ ) µ( ξ ) dξ =
™ G( s) µ ( t ’ s) ds.
™ (5.89)
ξ =’∞ s=’∞

The upper limit t is replaced by ∞. This is allowed because G( t) is de¬ned such
that G( t ’ ξ ) = 0 for ξ > t. After that we have substituted ξ = t ’ s. Substitution
of (5.88) into (5.89) leads to
87 5.4 Harmonic excitation of visco-elastic materials


G( s) e’iωs ds = iωµ0 G— ( ω) eiωt .
F( t) = iωµ0 e iωt
(5.90)
’∞

In this equation G— ( ω) can be recognized as the Fourier transform of G( t) (see
Appendix 5.5). Apparently the force has the same form as the strain, only the
force has a complex amplitude. If we de¬ne the Complex Modulus E— ( ω) as

E— ( ω) = iωG— ( ω) = E1 ( ω) + iE2 ( ω) , (5.91)

substitution of (5.91) into (5.90) gives the real part of F( t):

F( t) = µ0 E1 cos( ωt) ’ µ0 E2 sin( ωt) . (5.92)

It is clear again that E1 and E2 are the Storage and Loss modulus. The above
expression speci¬es the form of the force output in the time domain. We can also
directly derive the relation between E( ω) and G( ω) by using a Fourier transform
of Eq. (5.89):

F — ( ω) = G— ( ω) iωµ— ( ω) = E— ( ω) µ— ( ω) . (5.93)

The Complex Modulus is similar to the transfer function in system theory. The
Storage Modulus is the real part of the transfer function, the Loss Modulus is the
imaginary part.
When experiments are performed to characterize visco-elastic, biological mate-
rials, the results are often presented in the form of either the Storage and the Loss
Modulus as a function of the excitation frequency, or by using the absolute value of
the Complex Modulus, in combination with the phase shift between input (strain)
and output (force), as a function of the frequency. In the case of linear visco-
elastic behaviour these properties give a good representation of the material (this
has to be tested ¬rst). As a second step often a model is proposed, based on a com-
bination of springs and dashpots, to ˜¬t™ on the given moduli. If this is possible, the
material behaviour can be described with a limited number of material parameters
(the properties of the springs and dashpots) and all possible selections of proper-
ties to describe the material under consideration can be derived from each other.
This will be demonstrated in the next subsection for a particular, but frequently
used model, the standard linear model.



5.4.3 The standard linear model

The standard linear model can be represented with one dashpot and two springs,
as shown in Fig. 5.14. The upper part is composed of a linear spring, the lower
part shows a Maxwell element.
Fibres: time-dependent behaviour
88

c2 F1

F

c1 F2


Figure 5.14
The 3-parameter standard linear visco-elastic model.


Similar to the procedure as used in Section 5.3.2 the total strain of the Maxwell
element is considered as an addition of the strain in the dashpot (µd ) and the strain
in the spring (µs = µ ’ µd ). The following relations can be proposed for variables
that determine the standard linear model:
F = F1 + F2
F1 = c2 µ
F2 = c· µ™d

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