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