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a FM sin( θ) ’b W ’ c W0 = 0.
From this equation it follows that
b W + c W0
FM = .
a sin( θ)
Hence, from the force balance in the x- and y-direction the joint reaction forces
can be computed immediately.
Suppose that
π
θ= , a = 0.1 [m], b = 0.25 [m], c = 0.6 [m],
10
and
W = 5 [N], W0 = 10 [N],
then
FM = 235 [N], FJx = 223 [N], FJy = ’58 [N].


Exercises

3.1 The position vectors of the points P, Q, R and S are given, respectively:
xP = ex + 3ey
xQ = 4ex + 2ey
Static equilibrium
48

xR = 3ex + ey
xS = ’ex ’ ey .


The force vector F 1 = 2ex acts on point Q. The force vector F 2 = ’2ex
acts on point S.



P
F1
Q
xQ
ey
R
xP
xR
xS
ex
S
F2




Calculate the resulting moment of the forces F 1 and F 2 with respect to the
points P and R.
3.2 Determine the reaction forces and moments on the beam construction,
experienced at point A due to the fully clamped ¬xation in point A for
both con¬gurations in the ¬gure.



b c
a
F
F
A
A
c a
b
ey

ex
ez




3.3 A person is lying on a board, which is supported at both ends. A vertical
reaction force VA is acting on the board at point A and passes through the
origin of the coordinate system. A vertical reaction force VB is acting at the
point located in xB = ex . It is known that the weight of the person is F
and the weight of the board is P. The vector that determines the centroid C
of the person is given by: xC = ±ex + βey . Determine the reaction forces
RA and RB as a function of , ±, β, F and P.
49 Exercises

ey xC
C



VB
ex
VA
P
F




3.4 A swimmer with a weight G is standing at the edge of a diving board. The
centre of gravity of the swimmer is just above the edge. The distances a
and b are known. Determine the reaction forces on the board at B and C,
where the board is supported by rollers and a hinge, respectively.



B C

G

b a



Wall bars with length and weight G are placed under an angle ± against a
3.5
wall. Two ways to model the supports are given in the ¬gure. Calculate the
reaction forces on the bar in the supporting points as a function of G and ±
for both cases.


2




G
2 G

ey
ey
± ±

ex ex
4 The mechanical behaviour of fibres


4.1 Introduction

Fibres and ¬bre-like structures play an important role in the mechanical proper-
ties of biological tissues. Fibre-like structures may be found in almost all human
tissues. A typical example is the ¬bre reinforcement in a heart valve, Fig. 4.1(a).
Another illustration is found in the intervertebral disc as shown in Fig. 4.1(b).
Fibre reinforcement, largely inspired by nature, is frequently used in prosthesis
design to optimize mechanical performance. An example is found in the aortic
valve prosthesis, see Fig. 4.2.
Fibres are long slender bodies and, essentially, have a tensile load bearing
capacity along the ¬bre direction only. The most simple approximation of the,
often complicated, mechanical behaviour of ¬bres is to assume that they behave
elastically. In that case ¬bres have much in common with springs. The objective of
this chapter is to formulate a relation between the force in the ¬bre and the change
in length of a ¬bre. Such a relation is called a constitutive model.




4.2 Elastic fibres in one dimension

Assume, for the time being, that the ¬bre is represented by a simple spring as
sketched in Fig. 4.3. At the left end the spring is attached to the wall while the right
end is loaded with a certain force F. If no load is applied to the spring (¬bre) the
length of the spring equals 0 , called the reference or initial length. After loading
of the spring the length changes to , called the current length. It is assumed that
there exists a linear relationship between the change in length of the ¬bre ’ 0
and the applied force:

F = a( ’ 0) . (4.1)

The constant a re¬‚ects the stiffness properties of the spring and can be identi¬ed
by, for instance, attaching a known weight, i.e. a known force, to the spring in
51 4.2 Elastic fibres in one dimension




(a) Fibres in a heart valve. (b) Fibres in the intervertebral disc
Courtesy Mrs A. Balguid

Figure 4.1
Examples of ¬bre structures.




Figure 4.2
Fibre reinforcement in a stented valve prosthesis [5].

0




F


Figure 4.3
Unloaded and loaded spring.



vertical position and measuring the extension of the spring. However, formulat-
ing the force-extension relation as in Eq. (4.1) is, although formally correct, not
very convenient. If another spring was considered with the same intrinsic proper-
ties, while the initial length was different, the coef¬cient a would change as well.
Therefore the relation represented by Eq. (4.1) is scaled with the initial, unloaded,
length:
’ 0
=c ’1 .
F=c (4.2)
0 0
The mechanical behaviour of fibres
52

The coef¬cient c is an intrinsic property of the spring that is independent of the
unloaded length of the spring. It is common practice to introduce the so-called
stretch parameter », de¬ned as

»= , (4.3)
0

such that

F = c( » ’ 1) . (4.4)

The quantity » ’ 1 is usually referred to as strain, and it measures the amount
of deformation of the spring. Without any elongation of the spring the stretch
satis¬es » = 1, while the strain equals zero. The above force-strain relation rep-
resents linear elastic behaviour, as depicted in Fig. 4.4(a). If after stretching the
¬bre returns to its original length the force equals zero and no energy has been
dissipated. Since the stretch » is a dimensionless quantity and the unit of force is
[N] (Newton), the constant c also has unit [N].
For relatively small stretches » the actual behaviour of many biological ¬bres
may indeed be approximated by a linear relation between force and stretch. How-
ever, if the stretch exceeds a certain value the force-extension behaviour usually
becomes non-linear. In fact, in many cases ¬bres have a ¬nite extensibility. If
the stretch » approaches a critical value, say »c , the force in the ¬bre increases
sharply. A typical example of such a behaviour is modelled using the following
expression for the force stretch relation:
c
( » ’ 1) .
F= (4.5)
»’1
1’ »c ’1
F
F




»c
1
1
1
» »
(a) (b)

Figure 4.4
Force stretch relation for linear and non-linear spring.
53 4.3 A simple one-dimensional model of a skeletal muscle

For small extensions (» ≈ 1) the nominator in this expression satis¬es
»’1
1’ ≈ 1, (4.6)
»c ’ 1
such that the behaviour is identical to the linear spring. If » approaches the crit-
ical stretch »c the force indeed increases rapidly with increasing stretch. This is
re¬‚ected in Fig. 4.4(b). The solid line represents the non-linear, ¬nite extensibility
curve according to Eq. (4.5), while the dashed line represents the linear behaviour
according to Eq. (4.4).


4.3 A simple one-dimensional model of a skeletal muscle

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