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R·δ
R·R 1+2
=
R·R

R·δ
= R·R 1+2 .
R·R

If ± is a small number, then 1 + 2± ≈ 1 + ±, hence if δ is suf¬ciently small the
current length may be approximated by
R·δ
≈ R·R 1+ . (4.30)
R·R
Using

= |R| = R · R, (4.31)
0

this can be rewritten as:
R·δ
= + . (4.32)
0
0
The stretch » may now be expressed as:
R·δ
=1+
»= . (4.33)
2
0 0
Recall that the force-stretch relation for a spring is given by:
F B = c( » ’ 1) a, (4.34)
where a denotes the vector of unit length pointing from point A to point B and
where F B is the force acting on point B.
The mechanical behaviour of fibres
64

For suf¬ciently small displacements of the end points this vector may be
approximated by:
R R
a≈ = = a0 . (4.35)
|R| 0

Consequently, the following expression of the force vector F B is obtained

R·δ R δ
FB = c = c a0 · a0 , (4.36)
2
0 0
0
a0
»’1

or written in terms of the end point displacements
a0 · ( uB ’ uA )
FB = c a0 . (4.37)
0
In this expression three parts may be recognized. First the stiffness of the spring,
c, second the amount of stretch in the direction of the spring, see Fig. 4.15, also
called ¬bre strain µ:
a0 · ( uB ’ uA )
µ =»’1= (4.38)
0
and third the orientation of the spring, represented by the unit vector a0 .

Example 4.4 An immediate consequence of the linearization process is that, if the displacement
difference uB ’ uA is normal to the ¬bre axis, i.e. ( uB ’ uA ) ·a0 = 0, the force in
the ¬bre equals zero. Two examples of this are given in Fig. 4.16.


uB “ uA




a0, |a0| = 1

a … (uB “ uA)

Figure 4.15
Measure of the elongation of the spring.


uA uA
uB uB

Figure 4.16
Examples of ( uB ’ uA ) · a0 = 0 leading to F = 0.
65 4.5 Small fibre stretches

F
ey
F
’F1 1 F1
ex
“F2
P “F1
F2
2
π π
4 4
F2
(a) Two ¬bre con¬guration (b) Free body diagram

Figure 4.17
Two ¬bre con¬guration and free body diagram of point P.




Example 4.5 Suppose that two ¬bres have been arranged according to Fig. 4.17(a). Both ¬bres
have the same unloaded length 0 and elastic property c. At point P the two ¬bres
have been connected. A force F is applied to this point. The free body diagram
with respect to point P is given in Fig. 4.17(b). The orientation of each of the ¬bres
in space is represented by the vector ai (i = 1, 2). These vectors have been chosen
such that they point from the supports to point P, hence

a1 = ex
1√
a2 = 2( ’ex + ey ) .
2
If uP = ux ex + uy ey denotes the displacement of point P, the force vectors F 1 and
F 2 are described by
c
F1 = ( a1 · uP ) a1
0
c
= ux ex .
0

and
c
F2 = ( a2 · uP ) a2
0
c1
= ( ’ux + uy ) ( ’ex + ey ) .
02

The requirement of force equilibrium at point P implies

F ’ F 1 ’ F 2 = 0.

With F = Fx ex + Fy ey it follows that in the x-direction:
c
Fx + ( ’3ux + 2uy ) = 0,
20
The mechanical behaviour of fibres
66

while in the y-direction:
c
Fy ’ ( ’ux + uy ) = 0.
0
This gives two equations from which the two unknowns (ux and uy ) can be solved.


Exercises

4.1 The length change of a muscle, with respect to the length 0 in the relaxed
state, can be written as δ = ’ 0 .
(a) Give an expression for the force F in the activated muscle as a
function of c, δ, 0 and c .
(b) What is the magnitude of the force F when δ = 0?
4.2 Give a sketch of the force as a function of the active muscle length in the
diagram below

F

c



2 2
c 0 c 0


4.3 In reality the force length equation for an activated muscle in Eq. (4.9) is
only valid in a very limited range of extension ratios. The force that a sar-
comere (and thus a skeletal muscle) can exert has a maximum, as depicted
in the ¬gure below. Sarcomere lengths at several interesting points in the
graph are depicted by Li , with i = 1, . . . , 5.


L2
100

80
Maximum
L3
tension
60
(%)
L2 L3 L4 L5
L1
40
L4
20
0
Actin
Myosin
l5
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Sarcomere length [µm] Z disk
67 Exercises

(a) Explain why the force decreases when the sarcomere length exceeds
L4.
(b) The force versus length relation between L1 and L2 can be described
exactly with Eq. (4.9). Determine the value of c in the case where the
maximum force as given in the graph is 100 [N].
4.4 A muscle-tendon complex is loaded with a force F. The combination of
the muscle-tendon complex can schematically be depicted as given in the
¬gure.

F
muscle tendon




The force versus length relation in the muscle can be written as
m
F =c ’1 .
m m
m
c

The force versus length relation in the tendon can be written as
t
F =c ’1 .
t t
t
0

Determine the total length change of the muscle-tendon complex as a result
of the load F.
4.5 A ¬bre is marked on two sides with small dots. The position of these
dots is measured in an unloaded reference con¬guration. From these
measurements it appears that the positions are given as


x0,A = ’ex + 3ey
x0,B = 2ex + 3ey .

The ¬bre moves and in the current (deformed) con¬guration the positions
of points A and B are measured again:


xA = 4ex + 2ey
xB = 8ex ’ ey .

The constant in the force versus extension relation is c = 300 [N]. Deter-
mine the force vectors F A and F B in the deformed con¬guration.
The mechanical behaviour of fibres
68

a0
FA

x0,B
x0,A
ex a
xA
ey

xB
FB


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