Civil Engineering Reference
In-Depth Information
10
4
10
4
s
−
1
,[0
00]
T
ω
3
=
2
.
70
×
(
2
.
70
×
)
.
00
.
31 0
.
59 0
.
81 0
.
95 1
.
s
−
1
,[0
10
4
10
4
00]
T
)
ω
6
=
7
.
84
×
(
8
.
10
×
)
.
00
.
81 0
.
95 0
.
31
−
0
.
59
−
1
.
E= 210
×
10
6
kN/m
2
r
= 7.83 t/m
3
A=6
×
10
−
4
m
2
0.3m
Figure 10.14
4. Repeat question 3 using rod elements. Program 10.1 is easily modified to analyse
axial vibrations of rods. Make the following changes:
a) In the declarations change
ndof=4
to
ndof=2
and
nodof=2
to
nodof=1
.
b) In the
elements 2:
loop delete the statements that assign values to
mm(2,2)=...
and
mm(4,4)=...
and then change
mm(3,3)
to
mm(2,2)
.
c) Change routine
beam km
to
rod km
.
The data read into
prop
will now be
ea
(instead of
ei
)and
rhoa
.
(Ans: Using 5 equal rod elements with
EA
10
5
10
−
3
=
1
.
26
×
and
ρA
=
4
.
70
×
answers exactly the same as for question 3.)
5. Using the modified program from the previous question determine the first two natural
frequencies and eigenvectors for the stepped bar shown in Figure 10.15.
(Ans:
10
4
s
−
1
,[0
00]
T
ω
1
=
2
.
46
×
.
00
.
50 0
.
82 0
.
95 1
.
10
4
s
−
1
,[0
00]
T
)
ω
2
=
5
.
94
×
.
00
.
68
−
0
.
08
−
0
.
74
−
1
.
E= 210
×
10
6
kN/m
2
r
= 7.83 t/m
3
A
1
=6.45
×
10
−
4
m
2
A
2
=3.23
×
10
−
4
m
2
0.254 m
0.127 m
Figure 10.15
6. A vital attribute of an element “stiffness matrix” is that it should possess the right
number of “rigid body modes”, that is zero eigenvalues of the stiffness matrix. Test
the 2-node elastic rod element stiffness matrix and prove that it has only one zero
eigenvalue, corresponding to one rigid body mode (translation). If the rod has length
L
, calculate its non-
zero eigenvalue and hence natural frequency of free vibration assuming both lumped
and consistent mass. Compare with the “exact” value of
, cross-sectional area
A
, modulus
E
and mass per unit length
ρ
π/L
√
E/ρ
.
, Consistent: 2
√
3
/L
√
E/ρ
/L
√
E/ρ
(Ans: Lumped:2
)
