Chapter 8
Problems and Solutions Section 8.1 (8.1 through 8.7)
8.1
Consider the oneelement model of a bar discussed in Section 8.1.
Calculate the
finite element of the bar for the case that it is free at both ends rather than
clamped.
Solution:
The finite element for a rod is derived in section 8.1.
Since
u
1
is not
restrained equations (8.7) and (8.11) are the finite element matrices.
8.2
Calculate the natural frequencies of the freefree bar of Problem 8.1.
To what
does the first natural frequency correspond?
How do these values compare with
the exact values obtained from methods of Chapter 6?
Solution:
K
=
ΕΑ
λ
1
1
1
1
Μ
=
ρΑλ
6
2
1
1
2
Μ
1
Κ
=
6
Ε
ρλ
2
1
1
1
1
λ
1,2
= 0,
12
E
r l
2
and the corresponding eigenvectors are
x
1
= 1
1
[
]
T
/
2
and
x
2
= 1
 1
[
]
T
/
2
Therefore,
ϖ
1
= 0,
w
2
=
12
E
r l
2
The first natural frequency corresponds to the rigid body mode, or pure
translation.
From the solution to problem 6.8,
ϖ
1
= 0,
w
2
=
p
2
E
rl
2
The first natural frequency is predicted exactly while the second is 10.2% high.
A point of interest is that, due to symmetry, the first mode of a clampedfree rod
of length
l
/2 has the same natural frequency as the second mode of a freefree
rod of length
l
.
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8.3
Consider the system of Figure P8.3, consisting of a spring connected to a
clampedfree bar.
Calculate the finite element model and discuss the accuracy of
the frequency prediction of this model by comparing it with the method of
Chapter 6.
Solution:
The finite element for the clampedfree rod is given by (8.14) as
ρ
Al
3
ú
ú
u
2
(
t
) +
EA
l
u
2
(
t
) = 0
The spring has the effect of adding stiffness
K
at
u
2
.
Thus,
ρ
Al
3
ú
ú
u
2
(
t
) +
EA
l
+
K
æ
è
ö
ø
u
2
(
t
) = 0
From (1.16)
ϖ
=
3(
Kl
+
EA
)
r Al
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 Winter '09
 Hill
 3 L, 1m, 2 L, αν, xö lø

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