524
14 Nonlinear Vibration
For the next time step, after a few iterations
¨
x 4 = 0.24, ˙
x 4 = 0.692 and x 4 = 0.110
The spring force F s4 = 0.110 × 40 = 4.4 < 5.
Let us assume ¨
x 5 = −1, which means
¨
x 4 = 0.24 − 1 = −0.76,
˙
x 5 = 0.692 +
1
2
(0.24 − 1)(0.1) = 0.654,
x 5 = 0.110 + 0.692 × 0.1 +
1
6
× (2 × 0.24 − 1)(0.1)
2
= 0.178,
and
x 4 = 0.178 − 0.110 = 0.068
The spring force = 40 × 0.178 = 7.12 > 5
The spring at yield point = 5, therefore
¨
x 4 = 5 − 5 = 0
We assume ¨
x 5 = 0.24, so that ¨
x 4 = 0
˙
x 5 = 0.692 +
1
2
(0.24 + 0.24) × 0.1 = 0.716
and
x 5 = 0.110 + 0.692 × 0.1 +
1
6
(2 × 0.24 + 0.24) (0.1)
2
= 0.1804
As the maximum F s = 5 corresponding to F sy
¨
x 4 = 5 − 5 = 0
So the assumed value of ¨
x 5 is correct.
The above steps can be repeated for other time instants. The pattern of the response
curve expected is shown in Fig. 14.4.
The method can be easily extended to MDF systems. Nonlinear vibration has
been discussed very briefly in this chapter. The main purpose is nothing beyond
mere introduction of the topic. A number of textbooks are available on the subject,
which may be consulted for further details [1–5].