Biomedical Engineering Reference
In-Depth Information
Integration of the equilibrium equation yields
du
dx
=−
c
EA
,
where
c
is an integration constant. Integration of this result yields
1
EA
qx
+
c
EA
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
EA
du
dx
1
2
EA
qx
2
u
=−
+
x
=
L
=−
qL
+
c
=
F
,
which implies that
c
=
F
+
qL
.
Consequently, the displacement
u
is given by
1
2
EA
qx
2
F
qL
EA
+
u
=−
+
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
E
1
,
A
1
and
E
2
,
A
2
for the left and right bar, respectively. The length of the left bar equals
L
1
while
the right bar has length
L
2
. The boundary conditions are as depicted in the figure:
fixation at both
x
=
0 and
x
=
L
1
+
L
2
. Besides a concentrated force
F
is applied
at
x
=
L
1
. Our goal is to determine the displacement of the point
x
=
L
1
.
To solve this problem, we recognize two difficulties:
•
We have to find 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 first bar is given by
d
dx
E
1
A
1
du
1
dx
=
0f r 0
<
x
<
L
1
u
1
=
0 t
x
=
0
E
1
A
1
du
1
dx
=
N
1
at
x
=
L
1
.
