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In-Depth Information
The differentialequation of motion of the axially vibrating bar is
E
u
u
=
where
u
(
x
represents the mass density and
E
is the
modulusofelasticity. The boundary conditions are
u
(0
,
t
) is the axial displacement,
ρ
u
(
L
,
t
)
=
,
t
)
=
0.Letting
u
(
x
,
t
)
=
y
(
x
)sin
ω
t
, weobtain
E
y
y
=−
ω
2
y
(
L
)
y
(0)
=
=
0
The corresponding finite difference equations are
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
2
−
1
00
···
0
y
1
y
2
y
3
.
y
n
−
1
y
n
y
1
y
2
y
3
.
y
n
−
1
y
n
/
−
12
−
1
0
···
0
ω
2
0
−
12
−
1
···
0
E
L
n
=
.
.
.
.
.
.
.
.
.
.
.
.
···
−
−
00
12
1
00
···
0
−
11
2
(a) If the standard form of these equations is
Hz
=
λ
z
, write down
H
and the
transformationmatrix
P
in
y
=
Pz
. (b) Compute the lowest circular frequency of
the bar with
n
10, 100 and 10
00 u
tilizing the module
inversePower3
.
Note
: the
analytical solutionis
=
ω
1
=
π
√
E
/ρ/
(2
L
).
16.
u
P
1
2
n
- 1
n
P
x
k
L
The simply supported column is resting onanelasticfoundation of stiffness
k
(N/mpermeter length). An axialforce
P
acts on the column. The differential
equation and the boundary conditionsfor the lateral displacement
u
are
P
E I
u
+
k
E I
u
u
(4)
+
=
0
u
(0)
u
(
L
)
u
(0)
=
=
u
(
L
)
=
=
0
Using the mesh shown, the finite difference approximation of these equations is
+
α
)
u
1
−
4
u
2
+
u
3
=
λ
(2
u
1
−
(5
u
2
)
−
4
u
1
+
(6
+
α
)
u
2
−
4
u
3
+
u
4
=
λ
(
−
u
1
+
2
u
2
+
u
3
)
u
1
−
4
u
2
+
(6
+
α
)
u
3
−
4
u
4
+
u
5
=
λ
(
−
u
2
+
2
u
3
−
u
4
)
.
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