The ¬bres in the skeletal muscle have the unique capability to contract. On a

microscopic scale a muscle is composed of muscle ¬bres and myo¬brils. Myo¬b-

rils in turn are composed of actin and myosin proteins. The interaction of ¬laments

of these proteins through cross-bridges leads to the contractile properties of the

muscle.

The arrangement of these ¬laments into a sarcomere unit is sketched in

Fig. 4.5(a). Upon activation of the muscle the actin and myosin ¬laments move

with respect to each other causing the sarcomere to shorten. Upon de-activation

of the muscle the actin and myosin ¬laments return to their original positions

due to the elasticity of the surrounding tissue. In terms of modelling, the change

of the sarcomere length implies that the initial, unloaded length of the muscle

changes. Let 0 denote the length of the muscle in the non-activated state, while

c denotes the length of the muscle in the activated or contracted but unloaded

state, see Fig. 4.6. Now, in contrast with a simple elastic spring, the contracted

length c serves as the reference length, such that the force in the muscle may be

expressed as:

Z-disc

cross bridges

myosin

filaments

actin

filaments

sarcomere

(a) (b)

Figure 4.5

(a) Basic structure of a contractile element (sarcomere) of a muscle (b) Cross section of a muscle,

vertical stripes correspond to Z-discs.

The mechanical behaviour of fibres

54

0

unloaded, non-activated,

lc

unloaded, activated

F loaded, activated

Figure 4.6

Different reference and current lengths of a muscle.

F=c ’1 . (4.7)

c

For this it is assumed that, despite the contraction, c does not change. The acti-

vated, but unloaded, length c of the muscle may be expressed in terms of the non-

activated length 0 using a so-called activation or contraction stretch »c de¬ned as:

c

»c = . (4.8)

0

Typically »c < 1 since it represents a contractile action. For simplicity it is

assumed that »c is known for different degrees of activation of the muscle. Using

the activation stretch »c , the force-stretch relation for a muscle may be rewritten as

»

F=c ’1 with » = . (4.9)

»c 0

Effectively this expression implies that if the muscle is activated, represented by

a certain »c , and the muscle is not loaded, hence F = 0, the muscle will contract

such that

» = »c . (4.10)

If, on the other hand, the muscle is activated and forced to have constant length

0 , hence » = 1, the force in the muscle equals:

1

F=c ’1 . (4.11)

»c

Rather complicated models have been developed to describe the activation of

the muscle. A large group of models is based on experimental work by Hill [9]

and supply a phenomenological description of the non-linear activated muscle.

These models account for the effect of contraction velocity and for the difference

in activated and passive state of the muscle. Later, microstructural models were

developed, based on the sliding ¬lament theories of Huxley [12]. These models

55 4.4 Elastic fibres in three dimensions

can even account for the calcium activation of the muscle. However, a discussion

of these models is beyond the scope of this book.

4.4 Elastic fibres in three dimensions

The above one-dimensional force extension relation can be generalized to a

¬bre/spring having an arbitrary position in three-dimensional space. The locations

of the end points of the spring, say A and B, in the unstretched, initial con¬gura-

tion are denoted by x0,A and x0,B , respectively, see Fig. 4.7.

The initial length of the spring 0 follows from

= |x0,B ’ x0,A |. (4.12)

0

The initial orientation of the spring in space is denoted by the vector a0 having

unit length that follows from

x0,B ’ x0,A

a0 = . (4.13)

|x0,B ’ x0,A |

In the stretched, current con¬guration, the positions of the end points of the spring

are denoted by xA and xB . Therefore the current length of the spring can be

computed from

= |xB ’ xA |, (4.14)

while the current orientation in space of the spring may be characterized by the

vector a of unit length:

xB ’ x A

a= . (4.15)

|xB ’ xA |

a0

FA

x0,B

a

x0,A

xA

xB FB

Figure 4.7

Spring in three-dimensional space.

The mechanical behaviour of fibres

56

Clearly, in analogy with the scalar one-dimensional case, the stretch of the spring

» is de¬ned as

»= . (4.16)

0

To cause a stretch of the spring, a force must be applied to the end points A and B.

The forces applied on the end points are vectors represented by F A and F B . They

have equal magnitude but opposite direction:

F B = ’F A

and are parallel to the orientation vector a:

F B = Fa. (4.17)

The scalar F represents the magnitude of the force vector F B and for linearly

elastic springs this magnitude follows from the one-dimensional relation Eq. (4.4):

F = c( » ’ 1) . (4.18)

Therefore, the force vector acting on point B is given by

F B = c( » ’ 1) a, (4.19)

while the force vector acting on point A is given by

F A = ’c( » ’ 1) a. (4.20)

Example 4.1 Suppose a spring is mounted as depicted in Fig. 4.8. The spring is ¬xed in space at

point A while it is free to translate in the vertical direction at point B. A Cartesian

coordinate system is attached to point A, as depicted in Fig. 4.8. If point B is

moved in the vertical direction, the force on the spring at point B is computed

assuming linear elasticity according to Eq. (4.19). The length of the spring in the

ey ey y

ex

ex

A B A B

0

(a) (b)

Figure 4.8

Linear elastic spring, ¬xed at A and free to translate in the ey direction at point B. (a) undeformed

con¬guration (b) deformed con¬guration.

57 4.4 Elastic fibres in three dimensions

undeformed con¬guration is denoted by 0 . In the undeformed con¬guration the

force in the spring equals zero. In the current, deformed con¬guration the position

of point B follows from

xB = + yey .

0 ex

Point A is positioned at the origin:

xA = 0.

The force vector acting on the spring at point B is written as

F B = c( » ’ 1) a,

while the current length is written as

= |xB ’ xA | = + y2 .

2

0

The stretch » of the spring follows from

+ y2

2

0

»= = .

0 0

The orientation of the spring as represented by the unit vector a is given by

+ yey

xB ’ x A 0 ex

a= = .

|xB ’ xA | + y2

2

0

So, in conclusion the force vector F B applied to the spring at point B equals

⎛ ⎞

2 + y2

0 ex + yey

0

FB = c ⎝ ’ 1⎠ .

2 + y2

0

0

Given an initial length 0 = 10 [cm] and a spring constant c = 0.5 [N], the force

components of F B in the x- and y-direction (Fx and Fy , respectively) are repre-

sented in Fig. 4.9 as a function of the y-location of point B. Notice that both Fx

and Fy are non-linear functions of y even though the spring is linearly elastic.

The non-linearity stems from the fact that the stretch », is a non-linear function

of y. A non-linear response of this type is called geometrically non-linear since

it originates from geometrical effects rather than an intrinsic non-linear physical

response of the spring.

Example 4.2 Consider a spring, in the undeformed con¬guration, mounted at an angle ±0 with

respect to the x-axis, as depicted in Fig. 4.10. The spring is ¬xed in space at point

A while it is free to translate in the vertical direction at point B. A Cartesian

The mechanical behaviour of fibres

58

0.7

Fx [N]

0.6

Fy [N]

0.5

Force

0.4

0.3

0.2

0.1

0

0 5 10 15 20

y [cm]

Figure 4.9

Forces in the horizontal and vertical direction exerted on the spring in point B to displace point B in

the y-direction.

B

B

0

y

ey

ey y

±0 A

A

ex ex

cos(±0) cos(±0)

0 0