Biomedical Engineering Reference
In-Depth Information
Integration of the equilibrium equation yields
du
dx
=
c
EA
=
c
EA
0
(1
3
L
)
,
with
c
an integration constant that needs to be identified. Integration of this result
gives
x
+
EA
0
ln
1
+
3
L
3
cL
x
u
=
+
d
,
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
3
cL
EA
0
ln
(
1
+
0
)
+
d
=
d
,
u
(0)
=
0
=
hence
d
=
0. At
x
=
L
the force
F
is given, such that
x
=
L
=
EA
du
dx
c
=
F
.
Consequently
EA
0
ln
1
+
3
L
.
3
FL
x
u
=
The stress
σ
in the bar is computed directly from
σ
=
E
du
F
A
0
(1
dx
=
3
L
)
.
Example 6.4
Consider a bar of length
L
with constant Young's modulus
E
and cross section
A
. As before, the bar is fixed at
x
x
+
L
a force
F
is applied.
Furthermore, the bar is loaded by a constant distributed load
q
.
=
0 while at
x
=
x
q
F
L
x
=
0
Accordingly, the boundary value problem is given by
EA
du
dx
d
dx
=−
q
for 0
<
x
<
L
u
=
0 t
x
=
0
EA
du
dx
=
F
at
x
=
L
.
