Study of the Vibrations of a System Consisting …
63
3 Particular Solution of the System
In this study, we will consider that the shape is given by a function y = a n x
n , where
n ≥ 3. It results in y
= na n x
n−1 , y
= n(n − 1)x
n−2 , 1 +
y
2 = 1 + n
2 a
2
n x
2n−2 ,
h = a n L
n
10 . Taking into account that
L 10
0
P
2
(L 10 − x)dx =
P
4
L
2
10 ,
L 10
0
H (h − y)dx = H
na n
n + 1
L
n+1
10
(16)
from (14) and (15), it results
H =
n + 1
1 + n 2 a 2
n L
2n−2
10
E I
L
2
10
+
(n + 1)P
4na n L
n−1
10
(17)
and expressions (12) and (13) lead to
L 10 +
1
E A
L 10
0
H +
P
2
na n x
n−1
1 + n 2 a 2
n x 2n−2
dx −
L 10
0
1 + n 2 a 2
n x 2n−2 dx = 0.
(18)
Using the relation (17), (18) is a nonlinear one with the unknown a n . This equation
can be solved with the aid of numerical methods. In addition, the angle θ at the point
A, denoted by θ f , has the expression
sin θ f =
1
E I
H
na n
n + 1
L
n+1
10 −
P
4
L
2
10
.
(19)
4 Numerical Example
As example, we consider the following situation: two identical cantilever beams for
which the sections are squares of side a = 0.003 m, the lengths are L 10 = L 20 =
0.3 m. The excitation force is
P = P 0 + P 1 sin(ωt) + P 2 sin(2ωt) + P 3 sin(3ωt),
(20)
where P 0 = 50 N, P 1 = 25 N, P 2 = 10 N, P 3 = 5 N, ω = 20 s
−1 . The period of
simulation is t ∈ [0 . . . 5] s, while the step of simulation is selected t = 10
−3 s. The
integrals in (18) are calculated using the Simpson rule, the step of integration being
x = L 10
100. The elasticity moduli are E 1 = E 2 = E = 2 × 10
9 Pa. It results