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