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4.6 In the situation that is depicted in the drawing below only the gluteus
medius is active. During adduction of the bone the femur rotates. The
point of application of this muscle on the trochanter B follows a circular
path. The point of application on the acetabulum A is given by the vec-
tor xA = Ley . When the adduction angle φ = 0 the length of the muscle is
= c . Calculate the force F as a function of the adduction angle φ.


A

A
gluteus muscle
ey

Be
R x

φ
B™
5 Fibres: time-dependent behaviour


5.1 Introduction

In the previous chapter on ¬bres the material behaviour was constantly considered
to be elastic, meaning that a unique relation exists between the extensional force
and the deformation of the ¬bre. This implies that the force versus stretch curves
for the loading and unloading path are indentical. There is no history dependency
and all energy that is stored into the ¬bre during deformation is regained during
the unloading phase. This also implies that the rate of loading or unloading does
not affect the force versus stretch curves. However, most biological materials do
not behave elastically!
An example of a loading history and a typical response of a biological material
is shown in Figures 5.1(a) and (b). In Fig. 5.1(a) a deformation history is given
that might be used in an experiment to mechanically characterize some material
specimen. The specimen is stretched fast to a certain value, then the deformation
is ¬xed and after a certain time restored to zero. After a short resting period,
the stretch is applied again but to a higher value of the stretch. This deformation
cycle is repeated several times. In this case the length change is prescribed and the
associated force is measured. Fig. 5.1(b) shows the result of such a measurement.
When the length of the ¬bre is kept constant, the force decreases in time. This
phenomenon is called relaxation. Reversely, if a constant load is applied, the
length of the ¬bre will increase. This is called creep.
When the material is subjected to cyclic loading, the force versus stretch rela-
tion in the loading process is usually somewhat different from that in the unloading
process. This is called hysteresis and is demonstrated in Fig. 5.2. The differ-
ence in the response paths during loading and unloading implies that energy
is dissipated, usually in the form of heat, during the process. Most biological
materials show more or less the above given behaviour, which is called visco-
elastic behaviour. The present chapter discusses how to describe this behaviour
mathematically. Pure viscous behaviour, as can be attributed to an ideal ¬‚uid is
considered ¬rst. The description will be extended to linear visco-elastic behaviour,
Fibres: time-dependent behaviour
70

0.2

0.15




Strain [“]
0.1

0.05

0
0 500 1000 1500 2000 2500 3000
time [s]
(a)
400


200
Force [N]




0


“200
0 500 1000 1500 2000 2500 3000
time [s]
(b)

Figure 5.1
Loading history in a relaxation experiment. The deformation of the tissue specimen is prescribed.


400


300


200
Force [N]




100


0


“100


“200
0.05 0.1 0.15
0 0.2
Strain [“]

Figure 5.2
Force/strain curve for cyclic loading of a biological material.
71 5.2 Viscous behaviour

followed by a discussion on harmonic excitation, a technique that is often used to
determine material properties of visco-elastic materials.



5.2 Viscous behaviour

It is not surprising that biological tissues do not behave purely elastically, since a
large percentage of most tissues is water. The behaviour of water can be charac-
terized as ˜viscous™. Cast in a one-dimensional format, viscous behaviour during
elongation (as in a ¬bre) may be represented by
1d
F = c· , (5.1)
dt
where c· is the damping coef¬cient in [ Ns] and d /dt measures the rate of change
of the length of the ¬bre. Mechanically this force-elongational rate relation may
be represented by a dashpot (see Fig. 5.3). Generally:
1d
D= (5.2)
dt
is called the rate of deformation, which is related to the stretch parameter ». Recall
that

»= , (5.3)
0
such that
1d 1 d»
=
D= . (5.4)
» dt
dt
Ideally, a ¬‚uid stretching experiment should create a deformation pattern as visu-
alized in Fig. 5.4(a). In practice this is impossible, because the ¬‚uid has to be
spatially ¬xed and loaded, for instance via end plates, as depicted in Fig. 5.4(b). In
this experiment a ¬‚uid is placed between two parallel plates at an initial distance
0 . Next, the end plates are displaced and the force on the end plates is mea-
sured. A typical example of a stretched ¬lament is shown in Fig. 5.5. Although
this seems to be a simple experiment, it is rather dif¬cult to perform in practice.


l

F
F



Figure 5.3
Mechanical representation of a viscous ¬bre by means of a dashpot.
Fibres: time-dependent behaviour
72




R
l R
l


l0 l0



R0 R0
(a) Ideal elongation experiment (b) Actual elongational experiment
including end effects

Figure 5.4
Schematic representation of an elongation experiment for ¬‚uids.


time




Figure 5.5
Example of uniaxial testing experiment with a ¬‚uid.


x v



A B

Figure 5.6
Point B is moved with a constant velocity v.



Fig. 5.5 shows an experiment, where the ¬‚uid is a little extended initially, after
which gravitational sag continues the ¬lament stretching process. Ideally, a ¬la-
ment stretching experiment should be performed at a constant elongational rate.
This is not trivial to achieve. For instance, let one end of the dashpot, say point A
positioned at the origin (i.e. xA = 0), be ¬xed in space, while the other end, point
B is displaced with a constant velocity v, as depicted in Fig. 5.6. In that case the
position of point B is given by
73 5.2 Viscous behaviour

xB = + vt, (5.5)
0

with the initial length of the dashpot. The actual length at time t is given by:
0

= xB ’ xA = + vt. (5.6)
0

Hence, the elongational rate is given by
1d v
D= = . (5.7)
0 + vt
dt
This shows, that if one end is moved with a constant velocity, the elongational rate
decays with increasing time t. Maintaining a constant elongational rate is possible
if the velocity of point B is adjusted as a function of time. Indeed, a constant
elongational rate D implies that the length must satisfy
1d
= D, (5.8)
dt
subject to the initial condition = 0 at t = 0, while D is constant. Since
d ln ( ) 1d
= = D, (5.9)
dt dt
the solution of Eq. (5.8) is given by
= eDt . (5.10)
0

This means that to maintain a constant elongational rate, point B has to be
displaced exponentially in time, which is rather dif¬cult to achieve in practice
(moreover, try to imagine how the force can be measured during this type of
experiment!).


5.2.1 Small stretches: linearization

If uB and uA denote the end point displacements, introduce:
= u B ’ uA . (5.11)
The stretch » may be expressed as
+
0
»= . (5.12)
0
Introducing the strain µ as

µ= , (5.13)

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