Biomedical Engineering Reference
In-Depth Information
EA
du
dx
d
dx
+
q
=
0.
(6.20)
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
EA
du
dx
=
c
,
(6.21)
with
c
a constant. Nevertheless, the strain
ε
=
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
ε
(
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 specified. Two types of boundary
conditions are distinguished. Firstly,
essential
boundary conditions, formulated
in terms of specified boundary displacements. The displacement
u
(
x
) must at
least be specified 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-
stood as follows. Suppose that
u
satisfies the equilibrium equation Eq. (
6.20
).
Then, if the displacement
u
(
x
) is not specified at, at least, one end point, an
arbitrary constant displacement
c
may be added to
ˆ
ˆ
u
(
x
), while Eq. (
6.20
)for
this modified displacement field
u
(
x
)
ˆ
+
c
is still satisfied, since the strain is
given by
d
(
u
+
c
)
dx
du
dx
+
dc
dx
du
dx
ε
=
=
=
(6.22)
=
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
specified.
Secondly,
natural
boundary conditions, formulated in terms of external bound-
ary loads, may be specified, depending on the problem at hand. In the configura-
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)
