6-
27
Problems and Solutions Section 6.4 (6.30 through 6.39)
6.30
Calculate the first three natural frequencies of torsional vibration of a shaft of
Figure 6.7 clamped at
x
= 0, if a disk of inertia
J
0
= 10 kg m
2
/rad is attached to the
end of the shaft at
x
=
l
.
Assume that
l
= 0.5 m,
J
= 5 m
4
,
G
= 2.5
×
10
9
Pa,
ρ
=
2700 kg/m
3
.
Solution:
The equation of motion is
θ
ρ
θ
′
′
=
G
&
&
.
Assume separation of variables:
)
(
)
(
t
q
X
φ
θ
=
to get
2
or
σ
φ
φ
ρ
φ
ρ
φ
−
=
′
′
=
′
′
=
q
q
G
q
G
q
&
&
&
&
so that
˙˙
q
G
q
+
=
′′ +
=
ρ
σ
φ
σ φ
0
0
2
and
where
ω
ρ
σ
2
2
=
G
.
The clamped-inertia boundary condition is
θ
(0,t) = 0, and
GJ
).
,
(
)
,
(
0
t
l
J
t
l
θ
θ
&
&
=
′
This yields that
φ
(0) = 0 and
GJ
l q t
J
l q t
J
l
G
q t
′
=
=
φ
φ
φ
ρ
σ
( ) ( )
( )˙˙( )
( )
( )
0
0
2
or
J
l
J
l
′
=
φ
σ
ρ
φ
( )
( )
0
2
The solution of the spatial equation is of the form
φ
σ
σ
( )
sin
cos
x
A
x
B
x
=
+
but the clamped boundary condition yields
B
= 0.
The inertia boundary condition
yields
JA
l
J
A
l
l
J
J
l
l
l
σ
σ
σ
ρ
σ
σ
ρ
σ
σ
cos
sin
tan
)( .
)
=
=
=
0
2
0
1
5
0 5
m
10kg m
(2700kg/m
m
4
2
3
So the frequency equation is
tan
σ
σ
l
l
=
675
Using the MATLAB function
fsolve
; this has the solutions
:
σ
σ
σ
σ
σ
σ
1
2
3
1
2
3
1 5685
4 7054
7 8424
3 1462
9 4388
15 7312
l
l
l
=
=
=
=
=
=
.
.
.
.
.
.
or
Thus
ω
1
= 2134 rad/s
⇒
f
1
= 340 Hz
ω
2
= 6403 rad/s
⇒
f
2
= 1019 Hz
ω
3
= 10672 rad/s
⇒
f
3
= 1699 Hz

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