Biomedical Engineering Reference
In-Depth Information
N
2
E
2
A
2
L
2
.
u
2
(
L
1
)
=−
Because the bars must fit together at
x
=
L
1
we find
N
2
E
1
A
1
L
1
L
2
E
2
A
2
.
N
1
=−
Use of force equilibrium of the slice yields
N
2
E
1
A
1
L
1
L
2
E
2
A
2
+
N
2
+
F
=
0,
hence
−
F
N
2
=
.
E
1
A
1
L
1
L
2
E
2
A
2
1
+
Now
N
1
and
N
2
have been determined, it is possible to find an expression for
the displacement of the bar at point
x
L
1
as a function of the force
F
and the
material and geometrical properties of both bars:
=
FL
1
L
2
E
2
A
2
L
1
+
u
(
L
1
)
=
E
1
A
1
L
2
.
Exercises
6.1
A displacement field as a function of the coordinate
x
is given as:
u
(
x
)
=
ax
2
c
, with
a
,
b
and
c
constant coefficients. Determine the strain
field as a function of
x
.
+
bx
+
ax
2
6.2
A strain field as a function of coordinate
x
is given as
c
,
with
a
,
b
and
c
constant coefficients. Determine the displacement field
u
(
x
)
satisfying
u
(0)
ε
(
x
)
=
+
bx
+
=
0.
6.3
A bar with Young's modulus
E
and length
is clamped at
x
=
0 and loaded
with a force
F
at
x
=
. The cross section is a function of
x
(0
≤
x
≤
),
according to:
A
0
e
−
β
x
A
(
x
)
=
with
A
0
the cross section area at
x
=
0 and with
β
a positive constant.
F
x
(a) Determine the stress field
(
x
).
(b) Determine the displacement field
u
(
x
).
σ
