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0 2 4 6 8 10
time [s]

The only information that is available for the material is the result from
a creep test as given in the ¬gure below. In the creep test a tendon spec-
imen was loaded with a mass of 500 [kg]. The gravitational constant is
10 [m s’2 ].
Fibres: time-dependent behaviour

Result from creep test




total strain [’]






’5 0 5 10 15 20 25 30 35 40
time [s]

(a) We try to describe the material with a standard linear model. The
relaxation function for this material is:
G( t) = c2 + c1 e’t/„R .
Determine the value of c1 and c2 using the information in the ¬gure
with the result of the creep test.
(b) Sketch the force response on the strain history as given in the ¬rst
¬gure with the applied strain as a function of time.
6 Analysis of a one-dimensional
continuous elastic medium

6.1 Introduction

In the previous chapters the global behaviour of ¬bres was considered, without
much attention to the detailed shape of these structures. Only the length change of
the ¬bre played a role in the analysis. In the present chapter we address in a little
bit more detail the deformation of long slender structures. These can be tendons,
muscles, but also long bones. The aim is, to generalize the concepts introduced
in previous chapters for discrete systems (i.e. springs) to continuous systems. To
simplify matters the loading and deformation of a one-dimensional elastic bar is

6.2 Equilibrium in a subsection of a slender structure

Consider a straight bar as visualized in Fig. 6.1(a), loaded by an external force
F at x = L and ¬xed in space at x = 0. The ¬gure shows the bar with the
x-axis in the longitudinal or axial direction. It is assumed, that each cross section
initially perpendicular to the axis of the bar remains perpendicular to the axis after
loading. In fact it is assumed that all properties and displacements are a function
of the x-coordinate only.
The objective is, to compute the displacement of each cross section of the bar
due to the loading. The area of the cross section perpendicular to the central axis
of the bar as well as the mechanical properties of the bar may be a function of
the x-coordinate. Therefore the displacement may be a non-linear function of the
axial coordinate.
To compute the displacements of the cross sections a procedure analogous to
the discrete case (elastic springs) is followed. In contrast with the discrete case,
the equilibrium conditions are not applied on a global scale but are applied on a
local scale. For this purpose the free body diagram of an arbitrary slice of the bar,
for example the grey slice in Fig. 6.1(a), is investigated. This free body diagram
is depicted in Fig. 6.1(b). The left side of the slice is located at position x and the
Analysis of a one-dimensional continuous elastic medium


N (x + ”x)
N (x)


x + ”x
(a) Representation of a bar (b) Free body diagram of a slice at
position x

Figure 6.1
Bar and free body diagram of a slice of the bar.

slice has a length x. The net force on the left side of the slice equals N( x), while
on the right side of the slice a force N( x + x) is present. The net force on the
right side of the bar may be different from the net force on the left side of the slice
due to the presence of a so-called distributed volume force. A volume force Q is a
force per unit of volume, and may be due to, for instance, gravity. Integration over
the cross section area of the slice yields a load per unit length, called q.
If the distributed load q is assumed constant within the slice of thickness x,
force equilibrium of the slice implies that

N( x) = N( x + x) + q x. (6.1)

This may also be written as
N( x + x) ’ N( x)
+ q = 0. (6.2)
If the length of the slice x approaches zero, we can write
N( x + x) ’ N( x) dN
lim , (6.3)
x dx

where dN/dx denotes the derivative of N( x) with respect to x. The transition
expressed in Eq. (6.3) is illustrated in Fig. 6.2. In this graph a function N( x) is
sketched. The function N( x) is evaluated at x and x + x, while x is small.
When moving from x to x + x the function N( x) changes a small amount: from
N( x) to N( x + x). If x is suf¬ciently small the ratio N/ x de¬nes the tan-
gent line to the function N( x) at point x, and hence equals the derivative of N( x)
with respect to x. Notice that this implies that for suf¬ciently small x:
101 6.3 Stress and strain

N (x)

N (x + Dx)


x + Dx

Figure 6.2
First-order derivative of a function N( x).

N( x + x) ≈ N( x) +
x. (6.4)
The result of this transformation is that the force equilibrium relation Eq. (6.2)
may be written as
+ q = 0. (6.5)
If the load per unit length q equals zero, then the equilibrium equation reduces to
= 0, (6.6)
which means that the force N is constant throughout the bar. It actually has to be
equal to the force F applied to the right end of the bar see Fig. 6.1(a). Conse-
quently, if the slice of Fig. 6.1(b) is considered the force N( x) equals N( x + x).
In other words, the force in the bar can only be non-constant if q can be neglected.

6.3 Stress and strain

The equilibrium equation (6.5) derived in the previous section does not give infor-
mation about the deformation of the bar. For this purpose a relation between force
and strain or strain rate must be de¬ned, similar to the force-strain relation for an
elastic spring, discussed in Chapter 4. For continuous media it is more appropriate
to formulate a relation between force per unit area (stress) and a deformation mea-
sure, such as strain or strain rate. The concepts of stress and strain in continuous
media are introduced in this section.
In the one-dimensional case discussed in this chapter, the force N acting on
a cross section of the bar is assumed to be homogeneously distributed over the
Analysis of a one-dimensional continuous elastic medium


σ= N


(b) Stress σ if N is homogeneously
(a) Force acting on a surface
having area A distributed over the area A

Figure 6.3
Stress σ as a homogeneously distributed force N over an area A.





Figure 6.4
Inhomogeneous distribution of the force N.

surface of this cross section. If A denotes the area of the cross section, the stress
σ is de¬ned as
σ= . (6.7)
So, the stress is a force per unit area. If the force is not homogeneously distributed
over a surface, see Fig. 6.4, we must consider a small part of the surface, A.
Actually an in¬nitesimally small area A is considered. This surface area only
carries a part N of the total force N. Then, the stress σ is, formally, de¬ned as
σ = lim . (6.8)

This means that the stress is de¬ned by the ratio of an in¬nitesimal amount of
force N over an in¬nitesimal amount of area A, while the in¬nitesimal area
A approaches zero. In the following it is assumed that Eq. (6.7) can be applied.
103 6.3 Stress and strain

u (x)
u (x + Dx)


x + Dx


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