. 23
( 67 .)


ln (1 + 0) + d = d,
u( 0) = 0 =
hence d = 0. At x = L the force F is given, such that
= c = F.
dx x=L
3FL x
u= ln 1 + .
EA0 3L
The stress σ in the bar is computed directly from
du F
σ =E = x.
A0 ( 1 +
dx 3L )

Example 6.4 Consider a bar of length L with constant Young™s modulus E and cross section
A. As before, the bar is ¬xed at x = 0 while at x = L a force F is applied.
Furthermore, the bar is loaded by a constant distributed load q.

x q


Accordingly, the boundary value problem is given by
d du
= ’q for 0 < x < L
dx dx
u=0 at x = 0
=F at x = L.
109 6.5 Deformation of an inhomogeneous bar

Integration of the equilibrium equation yields
du 1 c
= ’ qx + ,
dx EA EA
where c is an integration constant. Integration of this result yields
1 c
u=’ qx2 + x + d.
Since at x = 0 the displacement u = 0, it follows immediately that d = 0. At
x = L the force F is prescribed, hence
= ’qL + c = F,
dx x=L
which implies that

c = F + qL.

Consequently, the displacement u is given by
F + qL
u=’ qx2 + x.
Example 6.5 Consider a system of two bars as depicted in Fig. 6.6(a). The Young™s modulus
and the cross section of both bars are constant and are given by E1 , A1 and E2 , A2
for the left and right bar, respectively. The length of the left bar equals L1 while
the right bar has length L2 . The boundary conditions are as depicted in the ¬gure:
¬xation at both x = 0 and x = L1 + L2 . Besides a concentrated force F is applied
at x = L1 . Our goal is to determine the displacement of the point x = L1 .

To solve this problem, we recognize two dif¬culties:
• We have to ¬nd a way to deal with the concentrated force F.
• The problem is statically indeterminate. Reaction forces cannot be uniquely determined
from force equilibrium alone.
The concentrated force has to be incorporated according to the following proce-
dure. Three free body diagrams have to be created by virtually cutting the bar just
left of the point where the force is applied and just right of that point. Thus three
free bodies can be distinguished: left bar, the right bar and a very thin slice around
the concentrated force. The free body diagrams are shown in Fig. 6.6(b).
The boundary value problem for the ¬rst bar is given by
= 0 for 0 < x < L1
E1 A1
dx dx
u1 = 0 at x = 0
= N1 at x = L1 .
E1 A1
Analysis of a one-dimensional continuous elastic medium


L1 L2


N1 N1 N2 N2
L1 L2

Figure 6.6
A two bar system and associated free body diagram.

Clearly, based on the previous examples, for the ¬rst bar the displacement ¬eld is
given by
u1 = x.
E1 A1
For the second bar the following boundary value problem holds:
d du2
= 0 for L1 < x < L1 + L2
E2 A2
dx dx
u2 = 0 at x = L1 + L2
= N2 at x = L1 .
E2 A2
The solution of this system is given by
N2 N2
u2 = x’ ( L1 + L2 ) .
E2 A2 E2 A2
We should realize that neither N1 nor N2 is known so far. However, there are two
additional equations that have to be satis¬ed. Force equilibrium of the slice (see
Fig. 6.6) requires that

’N1 + N2 + F = 0,

while the displacement ¬eld must be continuous at x = L1 : the two bars must
remain ¬tting together, hence

u1 ( L1 ) = u2 ( L1 ) . (6.25)

Based upon the solution for u1 and u2 it follows that
u1 ( L1 ) = L1 ,
E1 A1
111 Exercises

u2 ( L1 ) = ’ L2 .
E2 A2
Because the bars must ¬t together at x = L1 we ¬nd
E1 A1 L2
N1 = ’N2 .
L1 E2 A2
Use of force equilibrium of the slice yields
E1 A1 L2
+ N2 + F = 0,
L1 E2 A2
N2 = .
E1 A1 L2
1+ L1 E2 A2
Now N1 and N2 have been determined, it is possible to ¬nd an expression for
the displacement of the bar at point x = L1 as a function of the force F and the
material and geometrical properties of both bars:
FL1 L2
u( L1 ) = .
E2 A2 L1 + E1 A1 L2


A displacement ¬eld as a function of the coordinate x is given as: u(x) =
ax2 + bx + c, with a, b and c constant coef¬cients. Determine the strain
¬eld as a function of x.
A strain ¬eld as a function of coordinate x is given as µ(x) = ax2 + bx + c,
with a, b and c constant coef¬cients. Determine the displacement ¬eld u(x)
satisfying u(0) = 0.
A bar with Young™s modulus E and length is clamped at x = 0 and loaded
with a force F at x = . The cross section is a function of x (0 ¤ x ¤ ),
according to:
A( x) = A0 e’βx
with A0 the cross section area at x = 0 and with β a positive constant.



Determine the stress ¬eld σ (x).
(b) Determine the displacement ¬eld u(x).
Analysis of a one-dimensional continuous elastic medium

6.4 A bar with Young™s modulus E, cross section A and length is clamped
at x = 0. A distributed load q( x) is applied in x-direction on the bar, accor-
ding to:
q( x) = ± eβx ,


with ± and β constant load parameters.
(a) Determine the stress ¬eld σ (x).
(b) Determine the displacement ¬eld u(x).
A bar with Young™s modulus E, density ρ, cross section A and length
is hanging by its own weight. The gravitation acceleration is g. The point
where the bar is clamped is located at x = 0. Determine the displacement
¬eld u( x) and the length change of the bar, due to gravity.

x A


6.6 The central part of a femur is modelled as a straight tube with outer diame-
ter D, inner diameter d and length . Assume that the cortical bone behaves
like a linear elastic material with Young™s modulus E. In addition, assume
that the bone is loaded with an axial compressive force P.


x ød
113 Exercises

Determine the stress ¬eld σ (x).
Determine the displacement ¬eld u(x). Assume that at x = the
displacement u( ) satis¬es u( ) = 0.
6.7 Consider a muscle/tendon complex as shown in the ¬gure. To ¬nd out how
much the tendon and the muscle are extended when the complex as a whole
is loaded with a force F, a very crude two bar model can be used. The mus-
cle is modelled as a bar with length 1 , Young™s modulus E1 and cross
section A1 . At point B the muscle is attached to the tendon, which is mod-
elled as a second bar with length 2 , Young™s modulus E2 and cross section
A2 . At point A the muscle is attached to the bone, which we consider as
a rigid ¬xation. At point C a force F is applied in the direction of the bar

muscle tendon




. 23
( 67 .)