358
8 Free Vibration Analysis of Continuous Systems
Let us assume
φ(x) = C 1 x
2
+ C 2 x
3
Therefore,
dφ
dx
= 2C 1 x + 3C 2 x
2 and
d
2 φ
dx 2 = 2C 1 + 6C 2 x
Now,
k 11 =
L
0
E I · 2 · 2dx = 4E I · L
k 12 = k 21 =
L
0
E I · 2 · 6xdx = 6E I L
2
k 22 =
L
0
E I · 6x · 6xdx = 12E I L
3
m 11 =
L
0
ρ A x
2
· x
2 dx =
ρ A L
5
5
m 22 =
L
0
ρ A x
3
· x
3 dx =
ρ A L
7
7
m 12 = m 21 =
L
0
ρ A x
2
· x
3 dx =
ρ A L
6
6
Substituting the above values in Eq. (8.202), we get
−ρ A p
2
L
5
5
L
6
6
L
6
6
L
7
7
+
4 E I L 6 E I L
2
6 E I L
2 12 E I L
3
C 1
C 2
=
0
0
For a non-trivial solution of the above equation,
4 E I L − p
2
ρ A L
5
/5 6 E I L
2
− p
2
ρ A L
2
/6
6 E I L
2
− p
2
ρ A L
6
/6 12 E I L
3
− p
2
ρ A L
7
/7
= 0