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Figure 6.5
Displacements of a thin slice within a continuous bar.


Consider a slice of the bar having length x, as depicted in Fig. 6.5. The linear
strain µ is expressed in the stretch » of the slice by


µ = » ’ 1, (6.9)
where the stretch is the ratio of the deformed length of the slice and the initial
length. At position x the displacement of the cross section of the bar equals u( x),
while at x + x the displacement equals u( x + x). The initial length of the slice
equals:
= x, (6.10)
0

while the current length is given by
= x + u( x + x) ’ u( x) . (6.11)
Therefore, the stretch, that is the ratio of the deformed length over the initial length
is given by:
x + u( x + x) ’ u( x)
»= . (6.12)
x
Consequently, if the width of the slice x approaches zero, the strain is computed
from
x + u( x + x) ’ u( x)
µ = lim ’1
x
x’0
u( x + x) ’ u( x)
= lim . (6.13)
x
x’0
Using the de¬nition of the derivative this yields
du
µ=
. (6.14)
dx
In conclusion, the strain is de¬ned as the derivative of the displacement ¬eld u
with respect to the coordinate x.
Analysis of a one-dimensional continuous elastic medium
104

6.4 Elastic stress“strain relation

Recall, that the force“strain relation for an elastic spring at small, in¬nitesimal
displacements is given by
a·( uB ’ uA )
F=c a. (6.15)
l0
¬bre strain
Here, c represents the stiffness of the spring, while the unit vector a denotes the
orientation of the spring in space. In analogy with this, the (one-dimensional)
stress“strain relation for linearly elastic materials is de¬ned as
σ = Eµ, (6.16)

where E is the so-called Young™s modulus. Using the de¬nition of the strain in
terms of the derivative of the displacement ¬eld, this may also be written as
du
σ =E . (6.17)
dx
Example 6.1 For a given displacement ¬eld u( x), the stress ¬eld can be computed. Suppose, for
instance, that the Young™s modulus is constant and that u is given by a polynomial
expression, say:
u = a1 x + 2a2 x2 + 5a3 x3 ,
with a1 , a2 and a3 known coef¬cients. Then the stress will be
du
= E( a1 + 4a2 x + 15a3 x2 ) .
σ =E
dx


6.5 Deformation of an inhomogeneous bar

In case of a one-dimensional bar, the stress at each cross section is uniquely
de¬ned according to
N
σ=
. (6.18)
A
Substitution of N = Aσ into the equilibrium equation Eq. (6.5) yields
d( Aσ )
+ q = 0. (6.19)
dx
Subsequently, the stress“strain relation Eq. (6.17) is substituted such that the fol-
lowing second-order differential equation in terms of the displacement ¬eld u( x)
is obtained:
105 6.5 Deformation of an inhomogeneous bar

d du
+ q = 0.
EA (6.20)
dx dx

In the absence of a force per unit length, q = 0, the force in the bar must be
constant. The stress σ does not have to be constant, because the cross section area
A may be a function of the coordinate x.
Suppose that both the force and the stress are constant (this can only occur if
q = 0 and A is constant). Then it follows from Eq. (6.20) that

du
= c, (6.21)
EA
dx
with c a constant. Nevertheless, the strain µ = du/dx may be a function of the
coordinate x if the Young™s modulus is non-constant.
The solution of the differential Eq. (6.20) yields the displacement as a func-
tion of x, and once u( x) is known the strain µ( x) and the stress σ ( x) in the
bar can be retrieved. However, this differential equation can only be solved
if two appropriate boundary conditions are speci¬ed. Two types of boundary
conditions are distinguished. Firstly, essential boundary conditions, formulated
in terms of speci¬ed boundary displacements. The displacement u( x) must at
least be speci¬ed at one end point, and depending on the problem at hand,
possibly at two. This is required to uniquely determine u( x) and may be under-
ˆ
stood as follows. Suppose that u satis¬es the equilibrium equation Eq. (6.20).
Then, if the displacement u( x) is not speci¬ed at, at least, one end point, an
ˆ
arbitrary constant displacement c may be added to u( x), while Eq. (6.20) for
this modi¬ed displacement ¬eld u( x) +c is still satis¬ed, since the strain is
ˆ
given by

ˆ ˆ
d( u + c)
ˆ du dc du
= + =
µ= (6.22)
dx dx dx dx
=0

for any constant c. Such a constant c would correspond to a rigid body translation
of the bar. So, in conclusion, at least one essential boundary condition must be
speci¬ed.
Secondly, natural boundary conditions, formulated in terms of external bound-
ary loads, may be speci¬ed, depending on the problem at hand. In the con¬gura-
tion visualized in Fig. 6.1(a) the bar is loaded by an external load F at the right
end of the bar, at x = L, with L the length of the bar. At this boundary, the force
equals

N( x = L) = F = σ A. (6.23)
Analysis of a one-dimensional continuous elastic medium
106

Since σ = E du/dx, the natural boundary condition at x = L reads
du
= F.
EA (6.24)
dx
Because the equilibrium equation Eq. (6.20) is a second-order differential equa-
tion, two boundary conditions must be speci¬ed, one must be an essential bound-
ary condition (the displacement must be speci¬ed at least at one point to avoid
rigid body displacement) and the other may either be an essential or natural bound-
ary condition. The combination of the equilibrium equation with appropriate
boundary conditions is called a (determinate) boundary value problem.

Example 6.2 x
F


L
x=0


As a ¬rst example the solution of a well-de¬ned boundary value problem for a
homogeneous bar without a distributed load is analysed. Consider a bar of length
L that has a uniform cross section, hence A is constant, and with constant Young™s
modulus E. There is no volume load present, hence q = 0. At x = 0 the displace-
ment is suppressed, hence u = 0 at x = 0, while at the other end of the bar a force
F is applied. Then, the boundary value problem is fully described by the following
set of equations:
du
d
=0 0<x<L
EA for
dx dx
u = 0 at x=0
du
= F at x=L.
EA
dx
Integrating the equilibrium equation once yields
du
= c,
EA
dx
where c denotes an integration constant. This may also be written as
du c
= .
dx EA
Because both the Young™s modulus E and the cross section area A are constant,
integration of this relation gives
c
x + d,
u=
EA
with d yet another integration constant. So, the solution u( x) is known provided
that the integration constants c and d can be determined. For this purpose the
107 6.5 Deformation of an inhomogeneous bar

boundary conditions at both ends of the bar are used. First, since at x = 0 the
displacement u = 0, the integration constant d must be zero, hence
c
u= x.
EA
Second, at x = L the force is known, such that
du c
F = EA = EA = c.
dx EA
x=L
So, the (unique) solution to the boundary value problem reads
F
u= x.
EA
The strain µ is directly obtained via
du F
µ= = ,
dx EA
while the stress σ follows from
F
σ = Eµ =
A
as expected.

Example 6.3 Consider, as before, a bar of length L, clamped at one end and loaded by a force F
at the other end of the bar. The Young™s modulus E is constant throughout the bar,
but the cross section varies along the axis of the bar. Let the cross section A( x)
be given by
x
A = A0 1 + ,
3L
with A0 a constant, clearly representing the cross section area at x = 0.

x
F



L
x=0

The boundary value problem is de¬ned by the same set of equations as in the
previous example:
d du
=0 0<x<L
EA for
dx dx
u=0 x=0
at
du
=F x = L.
EA at
dx
Analysis of a one-dimensional continuous elastic medium
108

Integration of the equilibrium equation yields
du c c
= = x,
EA0 ( 1 +
dx EA 3L )
with c an integration constant that needs to be identi¬ed. Integration of this result
gives
3cL x
u= ln 1 + + d,
EA0 3L
with d another integration constant. The integration constants c and d can be deter-
mined by application of the boundary conditions at x = 0 and x = L. Since at
x = 0 the displacement u = 0 it follows that
3cL

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