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(a) (b)

Figure 4.10
(a) Linearly elastic spring in the undeformed con¬guration oriented at an angle ±0 with respect to
ex , ¬xed at A and free to translate in the ey -direction at point B (b) Spring in the deformed
con¬guration.



coordinate system is located at point A. If point B is moved in the vertical direc-
tion, the force in the spring is computed assuming linear elasticity according to
Eq. (4.19). The unstretched length of the spring is denoted by 0 such that the
current position of point B follows from
xB = 0 cos( ±0 ) ex + yey .
Point A is positioned at the origin:
xA = 0.
The current length of the spring satis¬es

= |xB ’ xA | = cos( ±0 ) )2 + y2 .
( 0
59 4.4 Elastic fibres in three dimensions

The force vector acting on the spring at point B is written as
F B = c( » ’ 1) a.
The stretch » of the spring follows from
cos( ±0 ) )2 + y2
( 0
»= = .
0 0
The orientation of the spring as represented by the unit vector a is given by
0 cos( ±0 ) ex + yey
xB ’ xA
a= = .
|xB ’ xA | 0 cos( ±0 ) ) + y2
2
(
So, in conclusion the force vector applied to the spring at point B, F B , equals

0 cos( ±0 ) ex + yey
0 cos( ±0 ) ) + y2
2
(
FB = c ’1 .
0 cos( ±0 +
) )2 y2
(
0

Given an initial length 0 = 1 [mm], a spring constant c = 0.5 [N] and an initial
orientation ±0 = π/4, the force components of F B in the x- and y-direction (Fx
and Fy , respectively) are represented in Fig. 4.11 as a function of the y-location
of point B. Notice that, as in the previous example, both Fx and Fy are non-linear
functions of y even though the spring is linearly elastic. It is remarkable to see that
with decreasing y, starting at the initial position y0 = 0 sin( ±0 ), the magnitude
of the force in the y-direction |Fy | ¬rst increases and thereafter decreases. This
demonstrates a so-called snap-through behaviour. If the translation of point B is


Fx [N] Fy [N]
0.1

T
R
0.05


0
Force




Q P
“0.05


“0.1


“0.15
y0
“1 “0.5 0 0.5 1
y [mm]

Figure 4.11
Forces in the horizontal and vertical direction exerted on the spring to displace point B in the
y-direction. Snap-through behaviour.
The mechanical behaviour of fibres
60

driven by an externally applied force, and point P is reached in the force versus
y-position curve, the y-coordinate of point B will suddenly move to point Q with
equal force magnitude. During the reverse path a snap through will occur from
point R to point T.


Example 4.3 The Achilles tendon is attached to the rear of the ankle (the calcaneus) and is
connected to two muscle groups: the gastrocnemius and the soleus, which, in turn,
are connected to the tibia, see Fig. 4.12(a). A schematic drawing of this, using a
lateral view, is given in Fig. 4.12(c). If the ankle is rotated with respect to the
pivot point O, i.e. the origin of the coordinate system, the attachment point A
is displaced causing a length change of the muscle system. The position of the
attachment point A is given by

xA = ’R sin( ±) ex ’ R cos( ±) ey ,

where R is the constant distance of the attachment point A to the pivot point. The
angle ± is de¬ned in clockwise direction. The muscles are connected to the tibia
at point B, hence:

xB = Hey ,

with H the distance of point B to the pivot point. The positions in the undeformed,
unstretched con¬guration of these points are

x0,A = ’R sin( ±0 ) ex ’ R cos( ±0 ) ey



B

1 tibia

H
1 2
ey
2
o ex
R
4
3
±
5 3
A
6
calcaneus
(a) The ankle and foot. (b) Ankle muscles, pos- (c) Location of muscle
(1) tibia, (2) ¬bula, terior view. (1) gastroc-
(3) medial malleolus, nemius, (2) soleus,
(4) lateral malleolus, (3) Achillles tendon
(5) talus, (6) calcaneus

Figure 4.12
Muscle attached to tibia and calcaneus.
61 4.5 Small fibre stretches

0
“0.005
“0.01
“0.015


F/c
“0.02
“0.025 Fx
Fy
“0.03
“0.035
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
±

Figure 4.13
Force in muscle.



and
x0,B = Hey .
Hence, the stretch of the muscle follows from

|xA ’ xB | ( R sin( ±) )2 + ( R cos( ±) + H)2
»= = ,
|x0,A ’ x0,B | ( R sin( ±0 ) )2 + ( R cos( ±0 ) + H)2

while the orientation of the muscle is given by

R sin( ±) ex + ( R cos( ±) + H) ey
a= .
( R sin( ±) )2 + ( R cos( ±) + H)2

From these results the force acting on the muscle at point B may be computed:
F B = c( » ’ 1) a.
The force components in the x- and y-direction, scaled by the constant c, are
depicted in Fig. 4.13 in case R = 5 [cm], H = 40 [cm] and an initial angle
±0 = π/4.



4.5 Small fibre stretches

As illustrated by the above example the ¬nite displacements of the end points
of a spring may cause a complicated non-linear response. In the limit of small
displacements of the end points a more manageable relation for the force in the
spring results. To arrive at the force versus displacement expression the concept
of displacement ¬rst needs to be formalized.
The mechanical behaviour of fibres
62

F

uB


u ’F uA x0,B

x0,A
x
x0




(a) Displacement (b) Small extension of the spring
vector u

Figure 4.14
Displacement vector and spring extension.




If, as before, the reference position of a certain point is denoted by x0 , and the
current position by x, then the displacement vector u of this point is de¬ned as the
difference between the current and initial position of the point, see Fig. 4.14(a):

u = x ’ x0 . (4.21)

If a spring, as in Fig. 4.14(b), is loaded by a force vector F, the end points A and
B will displace by the displacement vectors uA and uB , respectively. The current
position of the end points can be written as

xA = x0,A + uA
xB = x0,B + uB . (4.22)

The current length of the spring may also be expressed in terms of the end-point
displacements:

= |( x0,B + uB ) ’ ( x0,A + uA ) |. (4.23)

De¬ne R as the end-to-end vector from point A to point B in the initial, unloaded,
con¬guration, hence

R = x0,B ’ x0,A (4.24)

and let δ denote the difference in displacement of point B and point A:

δ = uB ’ uA . (4.25)

The current length can be expressed in terms of the vectors R and δ:

= |R + δ|. (4.26)
63 4.5 Small fibre stretches

This implies that (recall that a · a = |a||a| = |a|2 )
= ( R + δ) · ( R + δ)
2

= R · R + 2R · δ + δ · δ. (4.27)
If the end point displacements are suf¬ciently small, the inner product δ · δ will
be small compared to the other inner products in the above expression and may be
neglected. Therefore, to a good approximation we have
≈ R · R + 2R · δ.
2
(4.28)
Consequently, the current length may be written as

= R · R + 2R · δ. (4.29)
This may be rewritten in a more convenient form, bearing in mind that each of the
inner products yields a scalar:

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