61
Problems for Section 3.6 °Periodic but Not Harmonic Excitation
31. A machine is loaded by a periodic °sawtooth±shaped force, as depicted in Figure 3.50. The
load is assumed to have existed for a very long time. Model this force using: (a) a oneterm
Fourier series, (b) a threeterm Fourier series, and (c) a ²veterm Fourier series. In each case
solve for the response as a function of time. System properties are:
m
= 1
kg,
k
= 9
N/cm,
°
= 0
:
15
,
T
= 1
s, and
A
= 1
cm.
Figure 3.50: Sawtooth loading.
Solution:
First of all, we realize that this problem can be solved exactly by solving the problem
in discrete time intervals such that
m
°
x
1
+
c
_
x
1
+
kx
1
=
A
T
t
for
0
°
t
°
T
m
°
x
2
+
c
_
x
2
+
kx
2
=
A
T
t
±
A
for
T
°
t
°
2
T
.
.
.
The displacements and velocities are continuous at
t
=
nT
for
n
= 1
;
2
;
² ² ²
:
Here, we use the Fourier expansion for the periodic forcing.
The sawtooth function is given by
At=T;
0
°
t
°
T:
We ²rst have to evaluate the Fourier coe¢ cients as follows:
a
0
=
2
T
Z
T
0
At
T
dt
=
A
a
p
=
2
T
Z
T
0
At
T
cos
p!
T
tdt
=
2
T
2
A
cos
p!
T
T
+
p!
T
(sin
p!
T
T
)
T
±
1
p
2
!
2
T
b
p
=
2
T
Z
T
0
At
T
sin
p!
T
tdt
=
±
2
T
2
(
±
sin
p!
T
T
+
p!
T
(cos
p!
T
T
)
T
)
A
p
2
!
2
T
:
Note that the integration by part was used to evaluate
a
p
and
b
p
:
For example,
Z
T
0
t
{z}
u
cos
p!
T
tdt

{z
}
dv
=
t
{z}
u
sin
p!
T
t
p!
T

{z
}
v
°
°
°
°
°
°
°
°
T
0
±
Z
T
0
1
dt
{z}
du
sin
p!
T
t
p!
T

{z
}
v
:
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62
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
Since
!
T
= 2
±=T;
and
p
is an integer,
cos
p!
T
T
= cos 2
±p
= 1
;
and
sin
p!
T
T
= 0
:
Therefore,
a
p
= 0
and
b
p
=
±
A=p±:
The Fourier series representation of the force is then
F
(
t
) =
A
2
±
1
X
p
=1
A
p±
sin
p
2
±
T
t:
The one, three, and ²veterm expansions are respectively,
F
1
(
t
) =
A
2
±
A
±
sin
!
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 Fall '11
 Benaroya
 Fourier Series, Sin, total solution

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