14.2 Perturbation Method
523
Equations (4.12) and (4.13) give
˙
x 2 =
1
2
× 3 × 0.1 = 0.15
and
x 2 =
1
6
× (0.1)
2
× [2 × 3 + 3] = 0.015
Equation (a) now gives
¨
x 2 =
1.5
0.3
−
40 × 0.015
0.3
= 3
Therefore, calculation may be started for the next time step at t = 0.2 s.
Here, we introduce the incremental concept and use Eq. (14.3) as
¨
x i =
F i
m
− k i x i
(14.2b)
Let us assume ¨
x 2 = 1
¨
x 3 = ¨
x 2 + ¨
x 2 = 3 + 1 = 4
Therefore,
˙
x 3 = ˙
x 2 +
1
2
( ¨
x 2 + ¨
x 3 ))t = 0.15 +
1
2
(4 + 4) × 0.1 = 0.55
x 3 = 0.015 + 0.15 × 0.1 +
1
6
× (2 × 4 + 4) × (0.1)
2
= 0.05
x 2 = x 3 − x 2 = 0.05 − 0.015 = 0.035
F 2 = 3 − 1.5 = 1.5
and
¨
x 2 =
1.5
0.3
−
40 × 0.035
0.3
= 0.333
After a few iterations, it will be found that the values converge at ¨
x 2 = 0.6.
¨
x 3 = ¨
x 2 + ¨
x 2 = 3.6
˙
x 3 = 0.5
x 3 = 0.0475
The spring force F 3 = 0.0475 × 40 = 1.9 < 5.