(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: